Question

If $\dfrac{1}{{{1^2}}} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + \dfrac{1}{{{5^2}}} + ........ = \dfrac{{{\pi ^2}}}{{12}}$ then $\dfrac{1}{{{1^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{5^2}}} + ........\infty$is equal to?
A) $\dfrac{{{\pi ^2}}}{{12}}$
B) $\dfrac{{{\pi ^2}}}{{16}}$
C) $\dfrac{{{\pi ^2}}}{{18}}$
D) None of these

Answer 1

In first series the denominator has all the numbers odd and even but in second series asked there are only odd numbers in the denominator .It means even numbers are removed.

Complete step-by-step answer:
Given ,
$\dfrac{1}{{{1^2}}} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + \dfrac{1}{{{5^2}}} + ........ = \dfrac{{{\pi ^2}}}{{12}}$
Let , $S = \dfrac{1}{{{1^2}}} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + \dfrac{1}{{{5^2}}} + ........ = \dfrac{{{\pi ^2}}}{{12}}$
$S' = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{4^2}}} + \dfrac{1}{{{6^2}}}......$
$\Rightarrow S' = \sum\nolimits_{r = 1}^n {\dfrac{1}{{{{\left( {2r} \right)}^2}}}}$

$$\Rightarrow S' = \sum\nolimits_{r = 1}^n {\dfrac{1}{{4{{\left( r \right)}^2}}}} \\\ \Rightarrow S' = \dfrac{1}{4}\sum\nolimits_{r = 1}^n {\dfrac{1}{{{{\left( r \right)}^2}}}} \\\$$

The summation formed is equal to the given series.
$\Rightarrow S' = \dfrac{1}{4}S$
Now , to get sum of required series,

$$S - S' = S - \dfrac{1}{4}S \\\ \Rightarrow \dfrac{3}{4}S \\\ \Rightarrow \dfrac{3}{4} \times \dfrac{{{\pi ^2}}}{{12}} \\\ \Rightarrow \dfrac{{{\pi ^2}}}{{16}} \\\$$

So option B is the correct answer.

Additional information:
A series is a sum of a sequence of terms. That is, a series is a list of numbers with additional operations between them.

Note: Don’t start solving this type of example directly. First observe the sum here. Given a series is a combination of two series we can say.
So we splitted it into two parts. One series with even numbers only (S’). And thus series with odd numbers are automatically separated (required).