If ∑k=1∞1k2=π26 and si=∑k=1∞i(36k2−1)i, then 2(s1+s2)+1 is equal to

S1+S2=12[∑k=1∞{1(6k−1)2+1(6k+1)2}]⇒2(S1+S2)+1=112+152+172+1112+………(1) Let α=π26=112+122+132+……(2)α22=122+142+………(3)α32=132+162+……(4)α62=162+……−.−(5)Now, (2) –{(4)-(5) + (3) }gives2α3=2(S1+S2)+1⇒S1+S2=12[2α3−1]=π218−12

If ∑k=1∞1k2=π26 and si=∑k=1∞i(36k2−1)i, then 2(s1+s2)+1 is equal to

S1+S2=12[∑k=1∞{1(6k−1)2+1(6k+1)2}]⇒2(S1+S2)+1=112+152+172+1112+………(1) Let α=π26=112+122+132+……(2)α22=122+142+………(3)α32=132+162+……(4)α62=162+……−.−(5)Now, (2) –{(4)-(5) + (3) }gives2α3=2(S1+S2)+1⇒S1+S2=12[2α3−1]=π218−12

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If $\sum_{k = 1}^{\infty} \frac{1}{k^{2}} = \frac{π^{2}}{6}$ and $s_{i} = \sum_{k = 1}^{\infty} \frac{i}{{(36 k^{2} - 1)}^{i}}$ , then $2 (s_{1} + s_{2}) + 1$ is equal to

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$\frac{π^{2}}{9}$

$\frac{π^{2}}{18}$

$\frac{π}{9}$

$\frac{π^{2}}{18} - \frac{1}{2}$

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Detailed Solution

$S_{1} + S_{2} = \frac{1}{2} [\sum_{k = 1}^{\infty} {\frac{1}{{(6 k - 1)}^{2}} + \frac{1}{{(6 k + 1)}^{2}}}]$
$\Rightarrow 2 (S_{1} + S_{2}) + 1 = \frac{1}{1^{2}} + \frac{1}{5^{2}} + \frac{1}{7^{2}} + \frac{1}{11^{2}} + \dots \dots \dots$ (1)
$L e t α = \frac{π^{2}}{6} = \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots \dots$ (2)
$\frac{α}{2^{2}} = \frac{1}{2^{2}} + \frac{1}{4^{2}} + \dots \dots \dots$ (3)
$\frac{α}{3^{2}} = \frac{1}{3^{2}} + \frac{1}{6^{2}} + \dots \dots$ (4)
$\frac{α}{6^{2}} = \frac{1}{6^{2}} + \dots \dots - . - (5)$
Now, (2) –{(4)-(5) + (3) }gives
$\frac{2 α}{3} = 2 (S_{1} + S_{2}) + 1 \Rightarrow S_{1} + S_{2} = \frac{1}{2} [\frac{2 α}{3} - 1] = \frac{π^{2}}{18} - \frac{1}{2}$