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Quiz 10: Summing series

Last unanswered question  Question  Next unanswered question
 

Question 1

 
 
Given the series n ∑ ----1---- k=12k(2k +1) which of the statements below best describes it ?
a) The series is arithmetic   b) The series is geometric
c) The series is telescopic   d) None of the above

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
∑ k=1n----1---- 2k(2k+ 1) = -1- 2k ---1--- 2k + 1
= 1 2 -1 3 + 1 4 -1 5 + 1 6 -1 7 … - 1 2n-+1-,
so the series looks telescopic but the middle terms don’t cancel out.
 

Question 2

 
 
Given the series ∑n 3 +2(k - 1) k=1 , which of the statements below best describes it ?
a) The series is arithmetic   b) The series is geometric
c) The series is telescopic   d) None of the above

 

Your answer is correct.
∑n 3+ 2(k - 1) k=1 is arithmetic since it is of the form ∑n a + (k - 1)d k=1 .
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
 

Question 3

 
 
Given the series ∑n 3 2k-1 k=1 , which of the statements below best describes it ?
a) The series is arithmetic   b) The series is geometric
c) The series is telescopic   d) None of the above

 

Not correct. Choice (a) is false.
Your answer is correct.
n n ∑ --3- = ∑ 3(1)k-1 k=12k-1 k=1 2 which is of the form ∑n arn-1 k=1 .
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
 

Question 4

 
 
Find the sum of the series 9 ∑ 3- k=0 2k .
a) 1 6(1- (2)10)   b) 6(1- (1)9) 2
c) 3 (1 - 210) 2   d) None of the above

 

Your answer is correct.
∑9 3 1∑0 3 2k = 2k-1 k=0 k=1
= 3( 110 ) (21)-1-1 2 = 3( 110) 1-(21)- 2 = 6(1 - (12)10).
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
 

Question 5

 
 
Find the sum of the series ∑9 3+ 2k k=0
a) 99   b) 240
c) 198   d) 120

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
∑9 ∑10 3 + 2k = 3 + 2(k - 1) k=0 k=1
10(6-+9-×-2) = 2 = 120.
 

Question 6

 
 
Find the sum of the series ∑10 1 ------- k=1k(k + 1) .
a) 9 -- 10   b) 1 1 - (--)10 11
c) 10 -- 11   d) 1- ( 1-)10 10

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
∑ k=110k(1k+1) = ∑ k=1101k -k+11-
= 1 - 1-= 10 11 11 .
Not correct. Choice (d) is false.
 

Question 7

 
 
Write the following in sigma notation
2.3.4 + 3.4.5+ 4.5.6+ ⋅⋅⋅+ k(k + 1)(k + 2).
a) ∑k (i+ 1)(i+ 2)(i+ 3) i=1   b) ∑k i(i+ 1)(i+ 2) i=2
c) ∑k i(i+ 1)(i+ 2) i=1   d) k∑-1 (i+ 1)(i+ 2)(i+ 3) i=0

 

Not correct. Choice (a) is false.
Your answer is correct.
∑k i(i + 1)(i + 2) = 2.3.4 + 3.4.5 + ... + k(k + 1)(k + 2). i=2
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
 

Question 8

 
 
Find the sum of the series ∑5 ( 2 ) 2 + -k k=0 2 .
a) 191 16   b) 223 16
c) 255 16   d) 479 32

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
∑5 ( ) ∑5 ∑5 2+ 2k- = 2 + -1k-1 k=0 2 k=0 k=02 1 ∑5 1 = 12 + --1 + -k--1 2 k=1 2 31 255 = 14 + 16 = 16 .
Not correct. Choice (d) is false.
 

Question 9

 
 
$10000 is deposited in a bank account which pays 6% p.a. compounded every 6 months. The bank also charges $50 in fees every 6 months after calculating the interest. How much is in the account after 3 years ?
a) 11667.10   b) 11675.06
c) 11731.34   d) 11617.10

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
There are 6 periods of 6 months in 3 years earning 3%. At the end of the period we have
6 ∑6 i-1 A6 = 10000 × (1.03) - 50 (1.03) i=(1 ) = 10000 × (1.03)6 - 50 (1.03)6-- 1 0.03 = 11617.10
 

Question 10

 
 
Find the formula for the monthly repayments, R, on a 5 year $10000 car loan at a fixed rate of 15% p.a. compounded monthly.
a) 10000-×0.0125 R = 1- (1.0125)-60   b) 10000 × 0.0125 R = --(1.15)60---1-
c) -10000×-0.25 R = 1 - (1.25)-60   d) R = -10000×-0.125 1 - (1.125)-60

 

Your answer is correct.
The amount owing after n months is
( 1.0125n - 1 ) 10000(1.0125)n - R --0.125--- .
After 60 months there is nothing owing so
( ) 1.0125n---1 n R 0.125 = 10000(1.0125) .
After rearranging we get
10000×-0.0125 R = 1- (1.0125)-60.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.