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

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MathRevolution
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1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + = ^2/6. What is the value of 1/1^2 [#permalink] New post   29 Apr 2019, 23:48
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\(\frac{1}{1^2} + \frac{1}{2^2}+ \frac{1}{3^2} + \frac{1}{4^2} + … = \frac{π^2}{6}.\) What is the value of \(\frac{1}{1^2}+ \frac{1}{3^2} + \frac{1}{5^2} + … ?\)

A. \(\frac{π^2}{8}\)

B. \(\frac{π^2}{2}\)

C. \(π^2\)

D. \(2π^2\)

E. \(4π^2\)­
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ZeyadSerag
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Re: 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + = ^2/6. What is the value of 1/1^2 [#permalink] New post   30 Apr 2019, 05:33
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The first sequence is adding (1 / square of all integers)... The Second sequence is adding (1/square of all odd integers).

Clearly the second sequence should have a lower value than the first sequence.

Hence, going through the answers, only A has a lower value than (pi^2/6). Therefore, A is the correct answer.
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Re: 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + = ^2/6. What is the value of 1/1^2 [#permalink] New post   30 Apr 2019, 02:20
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MathRevolution wrote:
[GMAT math practice question]

\(\frac{1}{1^2} + \frac{1}{2^2}+ \frac{1}{3^2} + \frac{1}{4^2} + … = \frac{π^2}{6}.\) What is the value of \(\frac{1}{1^2}+ \frac{1}{3^2} + \frac{1}{5^2} + … ?\)

\(A. \frac{π^2}{8}\)
\(B. \frac{π^2}{2}\)
\(C. π^2\)
\(D. 2π^2\)
\(E. 4π^2\)


\(\frac{1}{1^2} + \frac{1}{2^2}+ \frac{1}{3^2} + \frac{1}{4^2} + … = \frac{π^2}{6}.\)
can also be written as:-


\(\frac{1}{1^2}+ \frac{1}{3^2} + \frac{1}{5^2} + …+\frac{1}{2^2}*(\frac{1}{1^2} + \frac{1}{2^2}+ \frac{1}{3^2} + \frac{1}{4^2} + …) = \frac{π^2}{6}.\)

\(\frac{1}{1^2}+ \frac{1}{3^2} + \frac{1}{5^2} + …+\frac{π^2}{6}*\frac{1}{2^2}= \frac{π^2}{6}.\)
solving- we will get \(\frac{π^2}{8}\)
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Re: 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + = ^2/6. What is the value of 1/1^2 [#permalink] New post   02 May 2019, 18:50
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Set \(x = \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + …\). Then

\(\frac{π^2}{6} =\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + …\)

\(= (\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + … ) + (\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + … )\)

\(= (\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + … ) + (\frac{1}{(1*2)^2} + \frac{1}{(2*2)^2} + \frac{1}{(2*3)^2} + … )\)

\(= (\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + … ) + (\frac{1}{4})(\frac{1}{1})^2 + \frac{1}{2^2} + \frac{1}{3^2} + … )\)

\(=x+(\frac{1}{4})(\frac{π^2}{6})\)

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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Re: 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + = ^2/6. What is the value of 1/1^2 [#permalink] New post   07 Sep 2020, 06:45
Here we could use a easier method, as we know the first sequence is more than the next sequence. The value of first sequence will be more than the second sequence.

And Hence, the value should be lesser than pi/6 which is pi/8.
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Re: 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + = ^2/6. What is the value of 1/1^2 [#permalink] New post   04 Mar 2021, 08:02
Instead of lengthy calculations, this can be answered by looking at the options and using a bit of logic.

(1/1^2)+(1/3^2)+(1/5^2)+... will always be less than (1/1^2)+(1/2^2)+(1/3^2)+(1/4^2)+... = (π^2/6)

Only option A is less than π^2/6.
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Re: 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + = ^2/6. What is the value of 1/1^2 [#permalink] New post   24 Mar 2024, 03:23
This is gonna be a stupid question, but I'm genuinely not aware of why:


why is it that the sequence with the squares of all integers is always less that the sequence with the squares of odd integers?
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Re: 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + = ^2/6. What is the value of 1/1^2 [#permalink] New post   24 Mar 2024, 03:31
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saynchalk wrote:
This is gonna be a stupid question, but I'm genuinely not aware of why:


why is it that the sequence with the squares of all integers is always less that the sequence with the squares of odd integers?


Because each term of the first sequence is more than (or equal to in the case of the first term) than each corresponding term of the second sequence.
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Re: 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + = ^2/6. What is the value of 1/1^2 [#permalink] New post   24 Mar 2024, 04:33
Bunuel wrote:
saynchalk wrote:
This is gonna be a stupid question, but I'm genuinely not aware of why:


why is it that the sequence with the squares of all integers is always less that the sequence with the squares of odd integers?

Because each term of the first sequence is more than (or equal to in the case of the first term) than each corresponding term of the second sequence.
 

I get it now, thanks Bunuel! :D
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Re: 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + = ^2/6. What is the value of 1/1^2 [#permalink]
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