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1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + … = π^2/6. What is the va

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by Max@Math Revolution » Mon Apr 29, 2019 10:42 pm

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C

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[GMAT math practice question]

1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ... = π^2/6. What is the value of 1/1^2 + 1/3^2 + 1/5^2 + ... ?

A. π^2/8
B. π^2/2
C. π^2
D. 2Ï€^2
E. 4Ï€^2
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by Max@Math Revolution » Thu May 02, 2019 5:38 pm
=>

Set x = 1/1^2 + 1/3^2 + 1/5^2 + ... . Then
Ï€^2/6 =1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ...
= (1/1^2 + 1/3^2 + 1/5^2 + ... ) + (1/2^2 + 1/4^2 + 1/6^2 + ... )
= (1/1^2 + 1/3^2 + 1/5^2 + ... ) + (1/(1*2)^2 + 1/(2*2)^2 + 1/(2*3)^2 + ... )
= (1/1^2 + 1/3^2 + 1/5^2 + ... ) + (1/4)(1/1)^2 + 1/2^2 + 1/3^2 + ... )
=x+(1/4)(Ï€^2/6)
So, x= π^2/6-(1/4)(π^2/6)=(3/4)(π^2/6)=π^2/8

Therefore, the answer is A.
Answer: A
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