If displaystyle∑k=1∞ (1/k2)=(π2/6) and Si= displaystyle∑k=1∞ (i/(36 k2-1)i), then S1+S2 is equal to

Question

If displaystyle∑k=1∞ (1/k2)=(π2/6) and Si= displaystyle∑k=1∞ (i/(36 k2-1)i), then S1+S2 is equal to (A) (π2/9)-(1/2) (B) (π2/12)+(1/2) (C) (π2/15)+(1/2) (D) (π2/18)-(1/2)

Tardigrade Admin · Accepted Answer

Correct answer is (d) (π2/18)-(1/2)

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

$\frac{π ^{2}}{9} - \frac{1}{2}$

$\frac{π ^{2}}{12} + \frac{1}{2}$

$\frac{π ^{2}}{15} + \frac{1}{2}$

$\frac{π ^{2}}{18} - \frac{1}{2}$

Correct answer is (d) $\frac{π ^{2}}{18} - \frac{1}{2}$

Q. If k=1∑∞​k21​=6π2​ and Si​=k=1∑∞​(36k2−1)ii​, then S1​+S2​ is equal to

9π2​−21​

12π2​+21​

15π2​+21​

18π2​−21​