# Exam #5 Sequences and Series

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1 College of the Redwoods Mathematics Department Math 30 College Algebra Exam #5 Sequences and Series David Arnold Don Hickethier Copyright c 000 Last Revision Date: April 9, 004 Version.00

2 Multiple Choice Questions Directions: In each of the following exercises, select the best answer and darken the corresponding oval on your scantron sheet.. Find the n th term of the sequence, 4, 9, 6,.... (a) a n = 3 (b) a n = n (c) a n = ( ) n (d) a n = ( ) n (n) (e) a n = n. Find the 5 th term of the sequence a n = (n ) 3. (a) 3 (b) 47 (c) 9 (d) 5 (e) Find the 5 th term of the recursive sequence where a =. a k+ = a k (a) 7 (b) 0 (c) 9 (d) 5 (e) 4. Find the 8 th term of the recursive sequence where a = and a = 3. a k+ = a k + a k (a) 34 (b) 55 (c) 99 (d) 0 (e) 9 5. Compute the sum 3 k= k. (a) 3 (b) 6 (c) 3 5 (d) (e) 6 6. Find the n th term of the arithmetic sequence 8, 5,,, 4,.... (a) a n = 8 + 3(n ) (b) a n = 8(3) n (c) a n = 8 3n (d) a n = 8 n (e) a n = 3 8(n ) 7. Find the two-hundredth term, a 00, of the sequence, 5, 8,,.... (a) 399 (b) 499 (c) 599 (d) 603 (e) 583

3 8. The first term of an arithmetic sequence is a = 3 and the eleventh term is a = 3. Find the n th term, a n, of the sequence. (a) a n = 3 + 3(n ) (b) a n = 3 + 0(n ) (c) a n = 3 + (n ) (d) a n = 3 + 0(n ) (e) a n = 3 + (n ) 9. Calculate the sum 00 k + 5. k= (a) 8,500 (b),700 (c) 9,00 (d) 38,500 (e) 4,00 0. Calculate the sum of the arithmetic series (a) 4 (b) 953 (c) 3906 (d) 7560 (e) Find the common ratio of the geometric sequence 4,,,... (a) (b) 4 (c) 4 (d) (e). The n th term of the geometric sequence 3, 6,, 4,... is (a) a n = 3( ) n (b) a n = ( 3) n (c) a n = (3) n (d) a n = 3( ) n (e) a n = 3() n 3. Find the sum of the first 8 terms of the geometric sequence,, 4, 8,... (a) 55 (b) 6 (c) 56 (d) 8 (e) 8 4. Compute the sum + a + a + a 3 + a 4 + a 5. (a) a6 (b) a6 a a (d) a5 (e) a a 5. Find the sum of the infinite series ( ) k 5. 3 (c) + a5 + a (a) 0 (b) 0 3 (c) 5 (d) 5 (e)

4 6. Find the sum of the infinite series + x + x + x 3 + given < x <. (a) xn x (b) (c) (d) x x x 7. Compute n C 3. (a) n! (b) n 6 3 n(n )(n ) (d) (e) n(n 3) 6 8. Solve for n. ( ) ( ) n + n = x (e) (c) n(n )(n ) (a) 9 (b) 8 (c) 7 (d) 0 (e) 6 9. Calculate the next row of Pascal s Triangle. 3 3 (a) (b) (c) (d) (e) Use the binomial theorem to expand (x y) 4. x (a) 6x 4 y 4 (b) 6x 4 3x 3 y + 4x y 8xy 3 + y 4 (c) 6x 4 8xy + y 4 (d) x 4 4x 3 y + 6x y 4xy 3 + y 4 (e) 6x 4 8x 3 y + x y 8xy 3 + y 4. Find the th term of (3x + y). (a) y (b) 44x y 9 (c) 56xy 0 (d) 33xy 0 (e) 4x y 9. One of the solutions of is 4 ( ) 4 (x) 4 k ( ) k = k (a) 0 (b) (c) (d) 4 (e) 3. Simplify (a) 5 (b) 7 (c) 9 (d) 30 (e) 3

5 Solutions to Multiple Choice Questions Solution to Question : The all of the terms of the sequence are perfect squares and the terms alternate sign. So the n th term of the sequence is a n = ( ) n (n). Solution to Question : To find the 5 th term of the sequence replace n with 5 in the sequence a n = (n ) 3. a 5 = (5 ) 3 = (4) 3 = 3 3 = 9 Solution to Question 3: The sequence is defined recursively by a k+ = a k. a = a = a = () = 3 a 3 = a = (3) = 5 a 4 = a 3 = (5) = 9 a 5 = a 4 = (9) = 7 Solution to Question 4: The sequence is defined recursively by a k+ = a k + a k with the first two terms given. Thus Solution to Question 5: 3 k= a = a = 3 a 3 = a + a = + 3 = 5 a 4 = a 3 + a = = 8 a 5 = a 4 + a 3 = = 3 a 6 = a 5 + a 4 = = a 7 = a 6 + a 5 = 3 + = 34 a 8 = a 7 + a 6 = + 34 = 55 k = = = 6 6 Solution to Question 6: The first term of the arithmetic series is a = 8 and the common difference is d = 3. Using the fact that for an arithmetic series the n th term is a n = a + (n )d, a n = 8 + 3(n )

6 Solution to Question 7: The sequence is arithmetic with first term a = 3 and common difference d = 3. Then a n = + (n )3 a n = + 3n 3 a n = 3n a 00 = 3(00) a 00 = 599 Solution to Question 8: Use the fact that for an arithmetic series the n th term is a n = a +(n )d to solve for d with a = 3 and a = 3. 3 = 3 + ( )d 0 = 0d = d Therefore, a n = 3 + (n ). Solution to Question 9: Write the series to notice the sum is an arithmetic series. 00 k + 5 = k= Indeed the sum is arithmetic with first term a = 7 and last term a 00 = 405. Now use the arithmetic series formula S n = n a + a n. 00 k + 5 = 00 ( ) = 00(4) = 4, 000 k= Solution to Question 0: The problem gives the first and last terms of the arithmetic series. To use the arithmetic series formula, S n = n (a + a n ), the number of terms needs to be determined. The first term is a = 3 and the common difference is d = 4, so a n = 3 + 4(n ). Thus Now it is possible to compute the sum. 3 = 3 + 4(n ) 0 = 4(n ) 30 = n 3 = n S 3 = 3 (3 + 3) = 3 (6) = 3(63) = 953

7 Solution to Question : The common ratio is r = a k+ a k = 4 = = Solution to Question : Use the n th term formula a n = a r n with a first term a = 3 and common ratio r = 6 3 = to get a n = 3( ) n. Solution to Question 3: The finite sum for a geometric series is S n = a ( r n ) where a r is the first term and r is the common ratio. The first term is a = and the common ratio is r = = 4 =. The sum of the first 8 terms is S 8 = ( 8 ) = 56 = 55 Solution to Question 4: The finite sum for a geometric series is S n = a ( r n ) r first term is a = and the common ratio is r = a = a. The sum of the 6 terms is where the S 6 = ( a6 ) a Solution to Question 5: The infinite sum for a geometric series is S = a r where a is the first term and r is the common ratio such that r <. The first term is a = 5 and the common ratio is r =. Since the common ratio is less than the infinite sum is 3 S = 5 /3 = 5 /3 = 5 Solution to Question 6: The infinite sum for a geometric series is S = a r where a is the first term and r is the common ratio such that r <. The first term is a = and the common ratio is r = x = x. Since < x < the infinite sum is S = x

8 Solution to Question 7: The number of combinations of n objects taken k at a time is nc k = n! (n k)!k!. So, nc 3 = Solution to Question 8: n! n(n )(n )(n 3)! n(n )(n ) = = (n 3)!3! 3!(n 3)! 6 ( ) ( ) n + n = (n + )! 5n! = 4!(n + 4)! 3!(n 3)! (n + )n! 5n! = 4(n 3)! 6(n 3)! n + 4 = 5 6 n + = 0 n = 9 Solution to Question 9: The next row of Pascal s Triangle is Solution to Question 0: The binomial expansion for (a + b) 4 is (a + b) 4 = a 4 + 4a 3 b + 6a b + 4ab 3 + b 4. To evaluate (x y) 4 let a = x and b = y in the equation above. (x y) 4 = (x) 4 + 4(x) 3 ( y) + 6(x) ( y) + 4(x)( y) 3 + ( y) 4 = 6x 4 3x 3 y + 4x y 8xy 3 + y 4 Solution to Question : The (k + ) st term in the binomial expansion of (a + b) n is ( n k) a n k b k. So the th term of the binomial expansion of (3x + y) is ( ) (3x) (y) 0 = (3x)y 0 = 33xy 0 0

9 Solution to Question : To solve 4 ( 4 k) (x) 4 k ( ) k = notice the left hand side of the equation is the fourth degree binomial expansion for (x ). That is 4 ( ) 4 (x) 4 k ( ) k = k (x ) 4 = (x ) = ± x = 3 x = Solution to Question 3: The sum resembles the sigma notation for the binomial expansion, (a + b) n = n ( n k) (a) n k (b) k with n = 5. If a = and b =, then = 5 = ( + ) 5 = 3 () n k () k k

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