Such formula by which any power of a binomial expression can be expanded in the form of a series is known as binomial theorem.

Transcription

1 Introduction To Binomial Theorem A Binomial Expression Any algebraic expression consisting of only two terms is known as a binomial expression. It's expansion in power of x is shown as the binomial expansion. For example: (i) a + x (ii) a 2 + 1/x 2 (iii) 4x - 6y Binomial Theorem Such formula by which any power of a binomial expression can be expanded in the form of a series is known as binomial theorem. It can be easily understood by examples. (a + x) 2 = a 2 + 2ax + x 2 (a + x) 2 = a 3 + 3a 2 x + 3ax 2 + x 3 Here, we see that the expression of (a + x) 2 is simple, we just multiply (a + x) by (a + x). Expansion of (a + x) 3 is little tougher, but what happens when the expansion is raised to the power of ten or more? So, we have to establish the formula for (a + x) n, where n is any integer. Let us define 'a' as the first term, 'x' as the second term and 'n' as the exponent. The total terms in the expansion of (a + x) 2 and (a + x) 3 are 3 and 4 respectively, which means that the number of terms in the expansion is one more than the exponent. So total number of terms in the expansion (a + x) n is (n + 1). Now, for n = 2 (a + x) 2 = a 2 x 0 + 2a 1 x 1 /1 + (2(2-1)/1*2)a 0 x 2 = (F.T.) n (S.T.) 0 + n/1! (F.T.) (n-1) (S.T.) 1 + (n(n-1)/2!) (F.T.) (n-2) (S.T.) 2 = a 2 + 2ax + x 2 Note: F.T. refers to first term i.e. 'a' and S.T. refers to second term i.e. 'x' Similarly, (a + x) 3 = a 3 + 3a 2 x + ((3(3-1)/(1*2))a 0 x 2 + ((3(3-1)(3-2))/(1*2*3))x 3 = a 3 + 3a 2 x + 3ax 2 + x 3 When n is a positive integer, then

2 (a + x) n = n C 0 a n + n C 1 a n-1 x + n C 2 a n-2 x n C r a n-r x r n C n x n, Where n C 0. n C 1. n C 2... n C n are called Binomial coefficients. Introduction To Binomial Theorem Proof of Binomial Theorem Proof of Binomial Theorem is very simple; we can prove it by using the mathematical induction. Proof: Step I: Let n = 1 L.H.S. = a + x R.H.S. = a + 1 C 1 x = a + x So, theorem is true for n = 1. Step II: Let the theorem be true for n = m, than P(m) : (a + x) m = m C 0 a m + m C 1 a m-1 x 1 + m C 2 a m m C m x m (i) Step III: We have to prove for n = m + 1 i.e. we have to prove that P(m + 1): (a + x) m+1 = m+1 C 0 a m+1 + m+1 C 1 a m x m+1 C m a x m + m+1 C m+1 x m+1...(ii) Multiplying by (a + x) on both sides in equation (i), we get, (a + x) m+1 = ( m C 0 a m+1 + m C 1 a m m C m-1 a 2 x m-1 + m C m a x m ) = ( m C 0 a m x + m C m 1 C 0 ) a m x m C m-1 a x m + m C m x m+1 ) Or, (a+x) m+1 = m C 0 a m+1 + ( m C 1 + m C 0 ) a m x ( m C m-1 + m C m-2 ) a 2 x m-1

3 + ( m C m + m C m-1 ) a x m + m C m x m+1 = m+1 C 0 a m+1 + m+1 C 1 a m x + m+1 C 2 a m-1 x m+1 C m-1 a 2 x m-1 + m+1 C m a x m + m+1 C m+1 x m+1. Hence, Proved. Expand (x + 1/x) 7. (x + 1/x)7 = 7 C 0 x C 1 x 6 (1/x) + 7 C 2 x 5 (1/x 2 ) + 7 C 3 x 4 + 1/x C 4 x 4 (1/x 3 ) + 7 C 5 x 2 + (1/x 5 ) + 7 C 6 x (1/x 6 ) + 7 C 7 (1/x 7 ) = x 7 + 7x x 3 + (35 x) + (35/x )+( 21/x 3 ) + (7/x 5 )+ (1/x 7 ). Properties Of Binomial Expansion 1. There are (n + 1) terms in the expansion of (a + b) n, the first and the last term being a n and b n respectively. If n C x = n C y, then either x = y or x + y = n => n C r = n C n-r = n!/r!(n-r)!. 2. The general term in the expansion of (a + x) n is (r + 1) th term given as T r+1 = n C r a n-r + x r. Similarly the general term in the expansion of (x + a) n is given as T r+1 = n C r x n-r a r. The terms are considered from the beginning. 3. The binomial coefficient in the expansion of (a + x) n which are equidistant from the beginning and the end are equal i.e. n C r = n C n-r.

4 Note: Here we are using n C r + n C r-1 = n+1 C r this concept will be discussed later in this chapter. Also, we have replace m C 0 by m+1 C 0 because numerical value of both is same i.e. 1. Similarly we replace m C m by m+1 C m+1. Find the expansion of (a-x) n. We know that (a + x) n = a n + na (n-1) x + (n(n-1)/2!) a (n-2) x x n. putting x = -x in the above expansion, we get, (a-x) n = [a + (-x)] n = a n + na (n-1) (-x) + (n(n-1)/2!) a (n-2) (-x) (-x) n. = a n - na (n-1) x + (n(n-1)/2!) a (n-2) x (-1) n x n. Find the value of (a+ (a 2-1)) 7 +(a- (a 2-1)) 7. Here, we have to find the sum of two expansions whose terms are numerically the same, but in the second expansion the second, fourth, sixth and eight terms are negative, and therefore cancel the corresponding terms of the first expansion. Hence, the given expression = 2 {a a 5 (a 2-1)+ 35 a 3 (a 2-1) 2 + 7a (a 2-1) 3 } = 2a (64 a a a 2-7)

5 Binomial Coefficients We know that, (a + x) n = n C 0 a n + n C 1 a n-1 x + n C 2 a n-2 x n C n x n... (i) Let us write equation (i) in particular form by putting a = 1. Now it becomes, (1 + x) n = n C 0 + n C 1 x + n C 2 x n C r x r n C n x n... (ii) Coefficients attached with different powers of x are called Binomial Coefficients. Now, again put a = 1 in the expression (a + x) n = a n + n a n-1 x + (n(n-1))/2! a n-2 x 2 + (n(n-1)(n-2))/3! a n-3 x (n(n-1)(n-2)...(n-r+1))/r! a n-r x r x n we get, x n (1+x) n = 1 + nx + (n(n-1))/2! x (n(n-1)(n-2)...(n-r+1))/r! x r (iii) If we compare coefficient of x r in expansions (ii) and (iii), we have, n C r = (n(n-1)(n-2)...(n-r+1))/r! = n!/r!(n-r)! What does it mean when we compare two expansions? Since expansion (ii) is valid for any value of x, we can replace x by 1/x, (x 0) in

6 it, we get, (1 + (1/x)) n = n C 0 + n C 1 1/x + n C 2 1/x n C r 1/x r n C n 1/x n Multiplying both sides by x n, it becomes, (1+x) n = n C 0 x n + n C 1 x n n C r x n-r n C n... (iv) Compare expansion (ii) and (iv). It will give us, that n C r = n C n-r, this is a beautiful result, which says that in the expansion of the equations of the like of (ii) the r th coefficient from the beginning is equal to the r th coefficient from the end. Sum Of Binomial Coefficients Putting x = 1 in the expansion (1+x) n = n C 0 + n C 1 x + n C 2 x n C x x n, we get, 2 n = n C 0 + n C 1 x + n C n C n. We kept x = 1, and got the desired result i.e. n r=0 C r = 2 n. Note: This one is very simple illustration of how we put some value of x and get the solution of the problem. It is very important how judiciously you exploit this property of binomial expansion. Find the value of C 0 + C 2 + C in the expansion of (1+x) n. We have, (1 + x) n = n C 0 + n C 1 x + n C 2 x n C n x n. Now put x = -x; (1-x) n = n C 0 - n C 1 x + n C 2 x (-1) n n C n x n. Now, adding both expansions, we get,

7 (1 + x) n + (1-x) n = 2[ n C 0 + n C 2 x 2 + n C 4 x ] Put x = 1 => (2 n +0)/2 = C 0 + C 2 + C or C 0 + C 2 + C = 2 n-1 We have found the sum of binomial coefficients. But if these coefficients are multiplied by some factors can we find the sum for such expressions? Yes, we can often find it by creatively applying what we have learnt. Let us see how differentiation and integration are useful in these situations. Suppose we want to calculate the value of n r=0 C r i.e. 0.C C 1 + 2C n n C n If we take a close look to the sum to be found, we find that coefficients are multiplied with respective powers of x. If we want to multiply the coefficient of x by its power differentiation is of help. Hence differentiate both sides of (1 + x) n = n C 0 + n C 1 x + n C 2 x 2 + n C 3 x n C n x n, with respect to x we get n(1 + x) n-1 = 1 C 1 x C 2 x n C n x n-1 Put x = 1, we get, n 2 n-1 = + 1 C C n C n. Or, n r=0 r C r = n 2 n-1, which is the answer. Now, think how to find the following sum C 0 /1 + C 1 /2 + C 2 / C n /n+1 =? In this sum coefficients are divided by the respective power of x + 1. This expression can be achieved by Integrating the expansion of (1 + x) n under proper limits.

8 Important: It has said earlier in this chapter that we should use and exploit the property that x can take any value in the expansion of (1 + x) n. Now, let us try to find the value of C 0 - C 2 + C 4 -C Analysing the above, expression, we find that C 0, C 2, C 4 are all coefficients of even powers of x. Had it been like C 0 + C 2 + C we could have evaluated simply by the method described earlier or had it been like C 0 -C 1 + C 2 -C we could have put x = -1 in the expansion of (1 + x) n and find the sum. But here the case is different. The expression consists of coefficients of only even powers. In such cases when there is alternative sign charge for the coefficient, of same nature of powers (even or odd) use of I (iota) comes to our rescue. (1 + x) n = C 0 + C 1 x + C 2 x (1 - x) n = C 0 - C 1 x + C 2 x By adding (1 + x) n + (1 - x) n = 2[C 0 + C 2 x 2 + C 4 x ]. Clearly use of -1 (iota) will generate the given expression. ((1+i) n + (1-i) n )/2 ] [C 0 - C 2 + C ] Coefficient Of A Particular Term

9 In the expression of (a + x) n, the coefficient of second term is n C 1, of the third term is n C 2, of he fourth term is n C 3 and so on. The suffix in each term being one less than the number of the term to which it applies; hence n C r is the coefficient of the (r+1) th term. This is called the General Term, because by giving different numerical values to r any of the coefficients may be found. Also, the indices of a and x in the (r + 1) th term are expressable in terms of r. Thus, General term of the expansion (a + x) n is dented as T r+1 = n C r a n-r x r = ((n(n-1)(n-2)...(n-r+1))/r!) a n-r x r. Note :General term is useful in finding the following in Binomial expression. (a) (b) Particular term. Middle term (c) Term independent of x. (d) Term containing the greatest coefficient of x. Find the fifth term in the expansion of (a + 2x 3 ) 17. General term of the expansion (a + x) n is T r+1 = n C r a n-r x r. Now, for the given expression, required term is T 5 = 17 C 4 a 13 (2x 3 ) 4. = a 13 x 12. Find the term independent of x in (1 + x + 2x 3 ) ((3/2)x 2 - (1/3x)) 9. We want to find the term independent of x in product of (1 + x + 2x 3 ) and

10 ((3/2)x 2 - (1/3x)) 9. So, we should find the terms containing x 0, x -1 and x -3 in the expansion ((3/2)x 2 - (1/3x)) 9 and multiply them respectively with the appropriate term from (1 + x + 2x 3 ) i.e. constant term i.e. 1, term containing x i.e. 1 and the term containing x 3 i.e. 2. Now, T r+1 = n C r a n-r x r [this is general term in (a+x) n ] = 9 C r ((3/2)x 2 ) 9-r (-(1/3x)) r This is general term in [((3/2)x 2 - (1/3x)) 9 ] (i) to find the term containing x 0 we use 18-3r = 0, i.e. r = 6 so coefficient of x 0 we use (-1) 6 9 C = 9 C 6 (1/ ). (ii) to find the term containing x -1 we use 18-3r = -1 i.e. r = 19/3 ( I) which is impossible. (iii) to find the term containing x -3 we put 18-3r = -3 i.e. r = 7. So, coefficient of x -3 in ((3/2)x 2 - (1/3x)) 9 = (-1) 7 9 C = - 9 C 7 (1/ ). Therefore, required coefficient = 9 C 6 (1/ ) C 7 (1/ ) = 17/54. Find the coefficient of x 24 in (x 2 +3a/x) 15. The general term ((r + 1)th term) is (x 2 +3a/x) 15 = 15 C r (x 2 ) 15-r (3a/x) r = 15 C r x 30-2r (3 r a r /x r ) = 15 C r 3 r a r x 30-3r If this term contains x 24, then 30-3r = 24 => 3r = 6 => r = 2. Therefore, the coefficient of x 24 = 15 C 2 9a 2. If the binomial co-efficient of the (2r + 4) th term and the (r - 2) th term in the expansion of (1 + x) 18 are equal, find the value r. The coefficient of (2r + 4) th term in (1 + x) 18 = 18 C 2r+3 and the coefficient of (r-

11 2) th term = 18 C r-3, so that 18 C 2r+3 = 18 C r-3. Either 2r + 3 = r - 3, => r = -6, Or 2r r - 3 = 18 => 3r = 18 => r = 6. (Since r ε N) If the 4 th term in the expansion (px + 1/x) n is independent of x, find the value of n. Also calculate p if the 4 th term is 5/2. Here T 4 = T 3+1 = n C 3 (px) 3 (1/x) n-3 = n C 3 p 3 x 6-n. T 4 is independent of x => 6 - n = 0 or n = 6. Now given T 4 = 5/2 => 6 C 3. p 3 = 5/2 => p 3 = 5/2. = 1/8 => p = 1/2. If the expansion of (1 + x) 43 the coefficient of (2r + 1) th term is equal to the coefficient of (r + 2) th term find r. Given in the expansion of (1 + x) 43 the coefficient of (2r + 1) th term = the coefficient of (r + 1) th term. 43 C 2r = 43 C r+1. Either 2r = r + 1 => r = 1, or 2r + r + 1 = 43 => r = 14. Hence r =1, 14 Find the coefficient of x 3 in the expansion of (1 + x + 2x 2 ) (2x 2 - (1/3x)) 9. T r+1 in (2x 2 - (1/3x)) 9 is 9 C r (2x 2 ) 9-r (-1/3x) r = 9 C r 2 9-r (-1/3) r x 18-3r => coefficient of x 3 is 9 C 5.24 (-1/3) 5 = -224/7.

12 Middle Term (i) When n is even Middle term of the expansion is the (n/2 + 1) th term i.e. n C n/2 a n/2 in the expansion of (a + b)n. (ii) When n is odd Middle terms of the expansion are the (n/2 + 1) th term and the ((n+3)/2) th term. These are given by, n C ((n-1)/2) a ((n+1)/2) a ((n-1)/2) and n C ((n+1)/2) a ((n-1)/2) a ((n+1)/2) expansion of (a + b)n. in the e.g. middle term in the expansion of (1+x) 4 and (1+x) 5. Expansion of (1+x) 4 have 5 terms, so third term is the middle term which is the ((4/2)+1) th term. Expansion of (1+x) 5 have 6 terms, so 3 rd and 4 th both are the middle terms, which are the ((5+1)/2) th and ((5+3)/2) th terms. Note: * r th term from the end = (n - r + 2) th term from the beginning. * If there are two middle terms, then the binomial co-efficients of two middle terms will be equal and those two co-efficients will be greatest. Find the middle term in the expansion of (1-2x + x 2 ) n. We have (1-2x + x 2 ) n = [(1 -x) 2 ] n = (1 - x) 2n. Here 2n is an even integer =>((2n/2) + 1) th i.e. (n+1) th term will be the middle term. Now (n+1) th term in (1-x) 2n = 2n C n (1) 2n-n (-x) n = 2n C n (-x) n = (2n)!/n!n! (-x) n. Greatest Binomial Coefficient

13 To determine the greatest coefficient in the binomial expansion, (1+x) n, when n is a positive integer. Coefficient of (Tr+1/Tr) = Cr/Cr-1 = (n-r+1)/r = ((n+1)/r) - 1. Now the (r+1) th binomial coefficient will be greater than the rth binomial coefficient when, T r+1 > T r => ((n+1)/r)-1 > 1 => (n+1)/2 >r.... (1) But r must be an integer, and therefore when n is even, the greatest binomial coefficient is given by the greatest value of r, consistent with (1) i.e., r = n/2 and hence the greatest binomial coefficient is n C n/2. Similarly in n be odd, the greatest binomial coefficient is given when, r = (n-1)/2 or (n+1)/2 and the coefficient itself will be n C (n+1)/2 or n C (n-1)/2, both being are equal Note: The greatest binomial coefficient is the binomial coefficient of the middle term. Show that the greatest the coefficient in the expansion of (x + 1/x) 2n is ( (2n-1).2 n )/n!. Since middle term has the greatest coefficient. So, greatest coefficient = coefficient of middle term = 2n C n = ( n )/n!n! = ( (2n-1).2 n )/n!. Numerically greatest term To determine the numerically greatest term in the expansion of (a + x) n, where n is a positive integer.

14 Consider Thus Note : {((n+1)/r) - 1} must be positive since n > r. Thus T r+1 will be the greatest term if, r has the greatest value as per the equation (1). Find the greatest term in the expansion of (3-2x) 9 when x = 1. T r+1 /T r = ((9-r+1)/r). (2x/3) >1 i.e. 20 > 5r If r = 4, then T r+1 = T r and these are the greatest terms. Thus 4 th and 5 th terms are numerically equal and greater than any other term and their value is equal C 3 (2/3) 3 = Find the greatest term in the expansion of (2 + 3x) 9 if x = 3/2. Here T r+1 /T r = ((n-r+1)/r)(3x/2) = ((10-r)/r)(3x/2), (where x = 3/2) = ((10-r)/r)(3x/2)(3/2) = ((10-r)/r).9/4 = (90-9r)/4r Therefore T r+1 > T r, if 90-9r > 4r.

15 => 90 > 13r => r < 90/13 and r being an integer, r = 6. Hence T r+1 = T 7 = T 6+1 = 9 C 6 (2) 3 (3x) 6 = /2. Given that the 4 th term in the expansion of (2 + (3/8)x) 10 has the maximum numerical value, find the range of values of x for which this will be true. Given 4 th term in (2 + (3/8)x) 10 = 2 10 (1 + (3/16)x) 10, is numerically greatest => T4/T3 > 1 and T5/T4 < 1 => => x > 2 and x < 64/21 => x ε [-(64/21),-2]U[2,(64/21)]. Find the greatest term in the expansion of 3(1 + (1/ 3)) 20. Let r th term be the greatest Since T r /T r+1 = (r/(21-r)). 3. Now T r /T r+1 > 1 => r > 21/( 3+1)... (1) Now again T r /T r+1 = (r/(21-r)) 3 < 1 => r < (22+ 3)/( 3+1)... (2) from (1) and (2) 22/( 3+1) < r < (22+ 3)/( 3+1) => r = 8 is the greatest term and its value is C 7 (1/ 3) 7 = 20 C 7 (1/27).

16 Particular Cases We have (a + x) n = a n + n C 1 a n-1 x + n C 2 a n-2 x r x n.... (1) (i) Putting x = -x in (1), we get (a-x) n = a n - n C 1 a n-1 x + n C 2 a n-2 x 2 - n C 2 a n-3 x (-1) r n C r a n-r x r (-1) n x n. (ii) Putting a = 1 in (1), we get, (1+x) n = n C 0 + n C 1 x + n C 2 x n C r x r n C n x n.... (A) (iii) Putting a = 1, x = -x in (1), we get (1-x) n = n C 0 - n C 1 x + n C 2 x 2 - n C 3 x (-1) r n C r x r +... (-1) n nc n x n... (B) Tips to Remember (a) T r /T r+1 = ((n-r+1)/r ).(x/a) for the binomial expansion of (a + x) n. (b) (n+1) C r = n C r + n C r-1. (c) r n C r = n n-1 C r-1 (d) (e) When n is even, (x + a) n + (x - a) n = 2(x n + n C 2 x n-2 a 2 + n C 4 x n-4 a n C n a n ). When n is odd, (x + a) n + (x - a) n = 2(x n + n C 2 x n-2 a n C n-1 x a n-1 ). When n is even (x + a) n - (x - a) n = 2( n C 1 x n-1 a + n C 3 x n-3 a n C n-1 x a n-1 ). When n is odd (x + a) n - (x - a) n = 2( n C 1 x n-1 a + n C 3 x n-3 a n C n a n ). Properties of Binomial Coefficients

17 For the sake of convenience, the coefficients n C o, n C 1,..., n C r,..., n C n are usually denoted by C o, C 1,..., C r,..., C n respectively Put x = 1 in (A) and get, 2n = C o + C C n... (D) Also putting x = -1 in (A) we get, 0 = C o - C 1 + C 2 - C => C 0 + C 2 + C C 1 + C 3 + C = 2 n. Hence C o + C 2 + C = C 1 + C 3 + C = 2 n-1. If (1 + x) n = C 0 + C 1 x + C 2 x C n x n, then prove that C 0 + (C o + C 1 ) + (C 0 + C 1 + C 2 ) +... (C 0 + C 1 + C C n-1 ) = n2 n-1 (where n is an even integer. C 0 + (C 0 + C 1 ) + (C 0 + C 1 + C 2 ) +... (C 0 + C 1 + C C n-1 ) = C 0 +(C 0 + C 1 + C C n-1 )+(C 0 + C 1 )+(C 0 + C 1 + C C n-2 )+... = (C 0 + C 1 + C C n )+ (C 0 + C 1 + C C n )+... n/2 times = (n/2)2 n = n. 2 n-1 Some Important Results 1. Differentiating (1+x) n = C 0 + C 1 x + C 2 x C n x n of both sides we have, n(1 + x) n-1 = C 1 + 2C 2 x + 3C 3 x n C n x n (E) Put x = 1 in (E) so that n2 n+1 = C 1 + 2C 2 + 3C nc n. Put x = -1 in (E) so that 0 = C 1-2C (-1) n-1 n C n. Differentiating (E) again and again we will have different results. 2. Integrating (1 + x) n, we have, ((1+x) n+1 )/(n+1) + C = C 0 x + C 1 x 2 /2 + C 2 x 3 / C n x n+1 /(n+1) is a constant) (where C

18 For x = 0, we get C = -1/(n+1). (F) Therefore ((1+x) n+1-1)/(n+1) = C 0 x + C 1 x 2 /2 + C 2 x 3 / C n x n+1 /(n+1)... Put x = 1 in (F) and get (2 n+1-1)/n+1= C 0 + C 1 /2 +...C n /n+1. Put x = -1 in (F) and get, 1/n+1 = C0 - C 1 /2 + C 2 / C n. Put x = 2 in (F) and get, (3 n+1-1)/n+1 = 2 C /2 C /3 C n+1 /n+1 = Problems Related to Series of Binomial Coefficients in Which Each Term is a Product of an Integer and a Binomial Coefficient, i.e. In the Form k. n C r. If (1+x)n = x r then prove that C 1 + 2C 2 + 3C nc n = n2 n-1. Method (i) : r th term of the given series r th term of the given series, t r = nc r => t r = r n/r n-1 C r-1 = n n-1 C r-1 (because n C r =n/r. n-1 C r-1 ) Sum of the series = Put x = 1 in the expansion of (1 + x) n-1, so that ( n-1 C 0 + n-1 C n-1 C n-1 ) = 2n-1 => = n.2n-1. Method (ii) : By Calculus We have (1 + x) n = C 0 + C 1 x + C 2 x C n x n.... (1) Differentiating (1) w.r.t. x, we get

19 n(1 + x) n-1 = C 1 + 2C 2 x + 3C 3 x n C n x n (2) Putting x = 1 in (2), we have, n 2 n-1 = C 1 + 2C nc n.... (3) If (1+ x) n = x r then prove that C C C (n+1)c n = 2n-1 (n+2). Method (i): r th term of the given series rth term of the given series t r = n C r-1 = [(r-1) + 1]. n C r-1 = (r-1) n C r-1 + n C r-1 = n. n-1 C r-2 + n C r-1 (because n C r-1 =n/(r-1). n-1 C r-2 ) Sum of the series = = n[ n-1 C 0 + n-1 C n-1 C n-1 ]+[ n C 0 + n C n C n ] = n.2 n-1 + 2n = 2 n-1 (n+2). Method (ii) by Calculus. We have (1 + x) n = C 0 + C 1 x + C 2 x C n x n.... (1) Multiplying (1) with x, we get x(1+x) n = C 0 x + C 1 x 2 + C 2 x C n x n (2) Differentiating (2) w.r.t. x, we have (1 + x) n + n(1 + x) n-1 x = C 0 x + 2C 1 x (n+1)c n x n... (3) Putting x = 1 in (3), we get 2n + n.2 n-1 = C 0 + 2C 1 + 3C (n+1)c n => C 0 + 2C 1 + 3C (n+1)c n = 2 n-1 (n+2).

20 Problems Related to Series of Binomial Coefficient in Which Each Term is Binomial Coefficient divided by an Integer, i.e. in the Form of n C r /k. If (1+x) n = Method (i) : r th term of the given series Method (ii): By Calculus (1+x) n = C 0 + C 1 x + C 2 x C n x n... (1) Integrating both the sides of (1) w.r.t. x between the limits 0 to x, we get... (2) Substituting x = 1 in (2), we get 2 n+1 /n+1 = C 0 + c 1 /2 + c 2 / c n/n+1. If (1+x)n = x r,show that Method I : r th term of the given series T r =

21 Method II : (By Calculus) (1 + x) n = C 0 + C 1 x + C 2 x C n x n => x(1+x) n = C 0 x + C 1 x 2 + C 2 x C n x n+1... (1) Integrating both the sides of (1) with respect to x Put x = 0, => k = 1/((n+1)(n+2)) Put x = 1, Problem Related to Series of Binomial Coefficients in Which Each Term is a Product of two Binomial Coefficients.. (a) If sum of lower suffices of binomial expansion in each term is the same i.e. n C 0 n C n + n C 1 n C n-1 + n C n n C n n C 0 i.e. 0 + n = 1 + (n-1) = 2 + (n-2) = n + 0. Then the series represents the coefficients of xn in the multiplication of the following two series

22 (1+x) n = C 0 + C 1 x + C 2 x C n x n and (1+x) n = C 0 + C 1 x + C 2 x C n x n. Prove that C 0 C r + C 1 C r+1 + C 2 C r C n-r C n = (2n)!/((n-r)!(n+r)!). We have, C 0 + C 1 x + C 2 x C n x n = (1+x) n... (1) Also C 0 x n + C 2 x n C n = (x+1) n... (2) Multiplying (1) and (2), we get (C 0 + C 1 x C n x n )(C 0 x n + C 1 x n-2 + C 2 x n C n ) = (1+x) 2n... (3) Equating coefficient of x n-r from both sides of (3), we get C 0 C r + C 1 C r+1 + C 2 C r C n-r C n = 2 n C n-r = (2n)/((n-r)!(n+r)!). Prove that C 2 o + C C 2 n = (2n)!/n!n!. Since (1 + x) n = C 0 + C 1 x + C 2 x C n x n,... (1) (x + 1) n = C 0 x n + C 1 x n C n,... (2) (C 0 + C 1 x + C 2 x C n x n )(C 0 x n + C 1 x n-1 + C 2 x n C n )=(1+x) 2n. Equating coefficient of x n, we get C C C C 2 n = 2n C n = (2n)!/n!n!.

23 If (1+x) n =, then prove that m C r n C 0 + m C r-1 n C 1 + m C r-2 n C m C 1 n C r-1 + m C 0 n C r = m+n C r where m, n, r are positive integers and r < m and r < n. (1+x) n = n C 0 + n C 1 x + n C 2 x n C r x r n C n x n... (1) and also (1+x) m = m C 0 + m C 1 x + m C 2 x m C r x r m C m x m... (2) Multiplying (1) and (2), we get ( n C 0 + n C 1 x + n C 2 x n C r x r n C n x n )x ( m C 0 + m C 1 x + m C 2 x m C r x r m C m x m ) = (1+x) m+n = m+n C 0 + m+2 C 1 x + m+n C 2 x m+n C r x r m+n C m+n x m+n Equating the coefficient of x r, we get m C r n C 0 + m C r-1 n C 1 + m C r-2 n C m C 1 n C r-1 + m C 0 n C r = m+n C r (b) If one series has constant lower suffices and other has varying lower suffices Prove that n C 0. 2n C n - n C 2n-2 1 C n + n C 2 / 2n-4 C n -...= 2 n. n C 0. 2n C n - n C 2n-2 1 C n + n C 2 / 2n-4 C n -... = coefficient of x n in[ n C 0 (1+x) 2n - n C 1 (1+x) 2n-2 + n C 2 (1+x) 2n ] = coefficient of x n in [ n C 0 ((1+x) 2 ) n - n C 1 ((1+x) 2 ) n-1 + n C 2 ((1+x) 2 ) n ]

24 = coefficient of x n in [(1+x) 2-1] n = coefficient of x n in (2x+x 2 ) n = co-efficient of x n in x n (2+x) n = 2 n. 1. Divisibility problems: Let (1 + x) n = 1 + n C 1 x + n C 2 x n C n x n. In any divisibility problem, we have to identify x and n. The number by which division is to be made can be, x, x 2 or x 3, but the number in the base is always expressed in form of (1+x). Find the remainder when is divided by = 7(50-1) 51 = 7( C C ) = 7( C C 50 50) = 7( C C 50 50) => remainder is Questions involving the greatest integer function. The questions generally involves working with binomial expression on surds. If I is the integral part and f is the fraction part of (2 + 3 ) n then prove that (l + f)(1- f) = 1. Also prove that I is an odd integer. (2 + 3) n, l + f where I is an integer and 0, f < 1. We have to show that I is odd and that (l + f)(1 - f) = 1 Here note that (2 + 3) n (2 + 3) n =(4-3) n = 1..(2 + 3) n (2 + 3) n =1. It is thus required to prove that (2 + 3) n =1-f

25 But, (2 + 3) n + (2 + 3) n = [2 n - C 1.2 n C 2 2 n-2.( 3) ] + [2 n - C 1.2 n C 2 2 n-2.( 3) ] =2[2 n - C 2.2 n C 4 2 n ] = even integer.. Now 0 < (2-3) < < (2-3) n < 1.. If (2-3)n = f' then l + f + f' = Even Now ) < f < 1 and 0 < f' < 1... (1).. f + f' = integer (1) and (2) imply that f + f' = 1(. 0 < f + f' < 2).. I is odd and f' = 1 - f => (I + f)(1 - f) = 1. Binomial expression with non- positive exponent The characteristics discussed hitherto were confined to positive integer n. If n takes any other value, then binomial theorem is written as: Say, we have to find out (x+a) n n I + If a > 1 then it is written as a n (a + (x/a)) n. This is expanded as a n (1+(x/n)) n = a n [1+n(x/a) + (n(n-1)/a) (x/a) 2 + (n(n-1)(n-2))/3! (x/a) ] Since n is not positive integer therefore the series on the right hand side will converge only for x/a < 1. Moreover, there are infinite terms in the expansion contrary to the binomial expansion for a positive integer n. Multinomial Expression If such a case arises, then it is not called Binomial Expansion, it is called Multinomial Expansion. If n ε N, then the general term of the multinomial expansion (x 1 + x 2 + x x k ) n is (n!/a 1!a 2!...a k!) (x 1 a1,x 2 a2,.. x k ak ), where a 1 + a 2 + a a k = n and a < a i < n, I = 1, 2, 3,... k and the total number of terms in the expansion is n+k-1 C n-1.

26 Solved Examples Example 1: Find the coefficient of the independent term of x in expansion of (3x - (2/x 2 )) 15. The general term of (3x - (2/x 2 )) 15 is written, as T r+1 = 15 C r (3x) 15-r (-2/x 2 ) r. It is independent of x if, 15 - r - 2r = 0 => r = 5.. T 6 = 15 C 5 (3) 10 (-2) 5 = - 16 C Example 2: If the coefficient of (2r + 4) th and (r - 2) th terms in the expansion of (1+x) 18 are equal then find the value of r. The general term of (1 + x) n is T r+1 = C r x r Hence coefficient of (2r + 4) th term will be T 2r+4 = T 2r+3+1 = 18 C 2r+3 and coefficient or (r - 2) th term will be T r-2 = T r-3+1 = 18 C r-3. => 18 C 2r+3 = 18 C r-3. => (2r + 3) + (r-3) = 18 (. n C r = n C K => r = k or r + k = n).. r = 6 Example 3: If a 1, a 2, a 3 and a 4 are the coefficients of any four consecutive terms in the expansion of (1+x) n then prove that:

27 a 1 /(a 1 +a 2 ) + a 2 /(a 3 +a 4 ) = 2a 2 /(a 2 +a 3 ) As a 1, a 2, a 3 and a 4 are coefficients of consecutive terms, then Let a 1 = n C r a 2 = n C r+1 a 3 = n C r+2 and a4 = n C r+3 Now a 1 /(a 1 +a 2 ) = n C r /( n C n r+ C r+1 ) = 1/(1+((n-r)/(r+1))) = (r+1)/(n+1) Similarly, a 2 /(a 2 +a 3 ) = (r+3)/(n+1) Now a 3 /(a 3 +a 4 ) + a 1 /(a 1 +a 2 ) = (2r+4)/(n+1) = 2(r+1)/(n+1) = 2a 2 /(a 2 +a 3 ) (Hence, proved) Example 4: Find out which one is larger or Let's try to find out in terms of remaining term i.e = (100+1) 50 - (100-1) 50 = (C C C ) = (C C C ) = 2[C C ] = 2[ C ] = [C ] > => >

28 Example 5: Find the value of the greatest term in the expansion of 3(1+(1/ 3)) 20. Let T r+1 be the greatest term, then T r < T r+1 > T r+2 Consider : T r+1 > T r => 20 C r (1/ 3) r > 20C r-1 (1/ 3) r-1 => ((20)!/(20-r)!r!) (1/( 3) r ) > ((20)!/(21-r)!(r-1)!) (1/( 3) r-1 ) => r < 21/( 3+1) => r < (i) Similarly, considering T r+1 > T r+2 => r > (ii) From (i) and (ii), we get r = 7 Hence greatest term = T 8 = 25840/9 Example 6: Find the coefficient of x 50 in the expansion of (1+x) x(1+x) x 2 (1+x) x Let S = (1 + x) x(1+x) x 999 (1+x) x 1000 This is an Arithmetic Geometric Series with r = x/(1+x) and d = 1. Now (x/(1+x)) S = x(1 + x) x 2 (1 + x) x x 1001 /(1+x) Subtracting we get, (1 - (x/(x+1))) S =(1+x) x(1+x) x x 1000 /(1+x)

29 or S = (1+x) x(1+x) x 2 (1+x) x 1000 (1+x)-1001x 1001 This is G.P. and sum is S = (1+x) x x 1001 So the coeff. of x 50 is = 1002 C 50 Example 7: Show that n C k (sin kx) cos (n-k)x = 2 n-1 sin(nx) We have n C k sin kx cos (n-k)x =1/2 n C k [sin (k x + nx - kx) + sin (kx - nx + kx)] =1/2 n C k sin n x + 1/2 n C k sin (2kx - nx) = 1/2 sin n x n C k 1/2 [ n C 0 sin (-nx) + n C 1 sin (2-n)x n C n-1 sin (n-2)x + n C n sin nx] = 2 n-1 sin nx + 0 (as terms in bracket, which are equidistant, from end and beginning will cancel each other). (Hence, proved) Example 8: If (15+6 6) 2n+1 = P, then prove that P(1 - F) = 9 2n+1 (where F is the fractional part of P). We can write P = (15+6 6) 2n+1 = I + F (Where I is integral and F is the fractional part of P) Let F' = (15+6 6) 2n+1

30 Note: 6 6 = => 0 < < 1 => 0 < (15+6 6) 2n+1 < 1 => 0 < F' < 1 Now, I + F = C 0 (15) 2n+1 + C 1 (15) 2n C 2 (15) 2n (6 6 ) F' = C 0 (15) 2n+1 - C 1 (15) 2n C 2 (15) 2n-1 (6 6 ) I + F + F' = 2[C 0 (15) 2n+1 + C 2 (15) 2n-1 (6 6 ) ] Term on R.H.S. is an even integer. => I + F + F' = Even integer => F + F' = Integer But, 0 < F < 1 and (F is fraction part) 0 < F' < 1 => 0 < F + F' < 2 Hence F + F' = 1 F' = (1-f).. P(1-F) = ( ) 2n+1 ( ) 2n+1 = (9) 2n+1 (Hence, proved) Example 9: Using 1 0 (tx+a-x) n dx,prove that 1 0 x k (1-x) (n-k) dx=[ n C k (n+1)] (-1) The given integral can easily be evaluated, as follows:

31 I = 0 1 (tx+a-x) n dx = 0 1 ((t-1)x+1) n dx = [((t-1)x+1) (n+1) /((n+1)(t-1))] 0 1 =(t (n+1) -1)/(n+1)(t-1)=1/(1+n) [1 + t + t t k t n ]... (i) Also, I = 0 1 (tx+(1-x)) n dx = 0 1 n C k (1-x) (n-k) t k x k dx... (ii) Comparing the coefficient of t k from (i) and (ii), we get, 0 1 n C k. x k (1-x) n-k dx=1/(n+1) => 0 1 x k (1-x) n-k dx = 1/( n C k (n+1)) (Hence, proved) Example 10: Prove that (-3) r-1 3n C 2r-1 = where k = 3n/2 and n is an even positive integer. Since n is even integer, let n = 2m, k = 3n/2 = 6m/2 =3m The summation becomes, S = Now, (1+x) 6m = 6m C 0 + 6m C 1 x + 6m C 2 x m C 6m-1 x 6m-1 + 6m C 6m x 6m... (i) (1-x) 6m = 6m C 0 + 6m C 1 (-x) + 6m C 2 (-x) m C 6m-1 (-x) 6m-1 + 6m C 6m (-x) 6m... (ii) => (1+x) 6m - (1-x) 6m = 2[ 6m C 1 x + 6m C 3 x m C 6m-1 x 6m-2 ]

32 ((1+x) 6m - (1-x) 6m )/2x = 6m C 1 + 6m C 3 x m C 6m-1 x 6m - 2 Let x 2 = y =>((1 + y)6m-(1- y)6m)/2 y= 6m C 1 + 6m C 3 y + 6m C 3 y+... 6m C 6m-1 y 3m-1 With y = -3, RHS becomes = S. LHS = (Hence, proved) Example 11: If k and n are two positive integers and S k = 1 k + 2 k n k Then show that m+ 1 C 1 S 1 + m+ 1 C 2 S m+ 1 C m S m = (1+n) m+1 - (1+n) We have, (1 + p) m+1 = m+1 C 0 + m+1 C 1 p + m+1 C 2 p 2 + m+1 C m+1 p m+1 Putting p = 1, 2, 3,..., n =>2 m+1-1 = m+1 C 0 + m+1 C 1 (1) + m+1 C 2 (1) m+1 C m (1) m 3 m+1-2 m+1 = m+1 C 0 + m+1 C 1 (2) + m+1 C 2 (2) m+1 C m (2) m 4 m+1-3 m+1 = m+1 C 0 + m+1 C 1 (3) + m+1 C 2 (3) m+1 C m (3) m

33 (1+n) m+1 -n m+1 = m+1 C 0 (n) + m+1 C 2 Σn + m+1 C 2 Σ(n) m+1 C m (n) m. Adding all these terms, we get (1+n) m+1-1 = m+1 C 0 (n) + m+1 C 1 S 1 + m+1 C 2 S m+1 C m S m => m+1 C 1 S 1 + m+1 C 2 S m+1 C m S m = (1+n) m+1 - (1+n) (Hence, proved) Example 12: Given that S n = 1 + q + q q n and P n = 1 + ((q+1)/2) + ((q+1)/2) ((q+1)/2) n show that n+1 C 1 + n+1 C 2 S 1 + n+1 C 2 S n+1 C n+1 S n = 2 n p n. Both S n and P n are geometric series S n = 1 + q + q q n = 1-q n-1 /1-q... (i) P n = 1 + ((q+1)/2) + ((q+1)/2) ((q+1)/2) n =(1-((q+1)/2) n+1 )/(1-(q+1)/2)=1/2 n (2 n+1 -(q+1) n+1 )/(1-q)... (ii) n+1 C r+1 S r = n+1 C r+1 ((1-q r+1 )/(1-q)) = n+1 C r+1 ((1/(1-q)) - (q r+1 /(1-q))) = (1/(1-q)) n+1 C r+1 - (1/(1-q)) n+1 C r+1 q r+1 => n+1 C r+1 S r =(1/(1-q)) n+1 C r+1 - (1/(1-q)) n+1 C r+1 q r+1 => n+1 C r+1 S r = 2 n p n (Hence, proved) Example 13: Show that if C r is the coefficient of x r in (1+x) n, then C r 3 = Coefficient of xn yn in the expression of ((1+x)(1+y)(x+y))n

34 (1+x) n = n C 0 + n C 1 x + n C 2 x n C n x n... (1) (1+y) n = n C 0 + n C 1 y + n C 2 y n C n y n... (2) (x+y) n = n C 0 x n + n C 1 x n-1 y + n C 2 x n n C n y n... (3) writing (1+y) n as (y+1) n and expanding it in decreasing powers of y. (y+1) n = n C 0 y n + n C 1 y n-1 + n C 2 y n n C n... (4) multiplying (1), (3) and (4) we get, (1+x) n (y+1) n (x+y) n = ( n C 0 + n C 1 x + n C 2 x n C n x n ) ( n C 0 y n + n C 1 y n n C n ) ( n C 0 x n + n C 1 x n-1 y n C n y n ) Now, coefficient of x n y n in RHS = n C n C n C n 3 C n = coefficient of x n y n in LHS = coefficient of x n y n in {(1+x)(1+y)(x+y)} n (Hence, proved) Example 14: If (1+x) n = C 0 + C 1 x + C 2 x C n x n (nεn) Then show that k 3 [C k /C k-1 ] = 1/12 (n)(n+1) 2 (n+2) C r /C r -1 = (n!/(n-r)!r!) (((r-1)!(n-r+1)!)/n!) = (((n+1)/r)-1) r 3 (C r /C r -1) = ((n+1)/r-1) r 3 = (n+1)r 2 - r 3 = (n+1)n(n+1)(2n+1)/6-(n 2 (n+1) 2 )/4

35 = n(n+1)/12 [4n 2 + 6n + 2-3n 2-3n] = n(n+1)(n 2 +3n+2)/12 n(n+1)(n+1)(n+2)/12=(n(n+1) 2 (n+2))/12 (Hence, proved) Example 15: If (1+x) n = C 0 + C 1 x + C 2 x C n x n then show that (i) 0 i<j n C i C j = 2 2n-1 - (ii) 0 i<j n (C i + C 2 j) = (n - 1) 2n C n + 2 2n (iii) 0 i<j n C i C j = n (2 2n-1-1/2 2n C n ) (i) We have (C 0 + C 1 + C C n ) 2 = C C C n i<j n C i C j we get, (2n) 2 = 2n C n i<j n C i C j therefore 0 i<j n C i C j = 2 2n-1 - (2n)!/2(n!) 2 (Hence, proved) (ii) 0 i<j n C i C j n(c C C C n2 ) i<j n C i C j = n 2n C n + 2{2 2n-1 - (2n)!/2(n!) 2 } [from part (i)] = n 2n C n + 2 2n - 2n C n = (n-1) 2n C n + 2 2n (Hence, proved) (iii) Let 0 i<j n C i C j replace i by (n - i) and j by (n - j), we have P = 0 i<j n (2n-i0j) C n-i C n-j = 0 i<j n (2n-(i+j)) C i C j [. n C n = n C n-r ]

36 = 2n 0 i<j n C i C j - P.. 0 i<j n (i+j) C i C j = n[2 2n-1 - (2n)!/2(n!) 2 ] (Hence, proved)