and S, =2,then S +S2 is equal to(36k-1)'k41n 1(B)12 2(A)(C)15(D)18 2112

Given that : If ∑k=1∞1k2=π26Si=∑k=1∞i(36k2-1)i In-order to find the value of S1+S2, S1 =…

Answered: If and S,= then S + S2 is equal to…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

and S, =2
,then S +S2 is equal to
(36k-1)'
k41
n 1
(B)
12 2
(A)
(C)
15
(D)
18 2
112

Expert Solution

Step 1

Given that :

If $\sum_{k = 1}^{\infty} \frac{1}{k^{2}} = \frac{π^{2}}{6} S_{i} = \sum_{k = 1}^{\infty} \frac{i}{{(36 k^{2} - 1)}^{i}}$

In-order to find the value of S₁+S₂,

$S_{1} = \sum_{k = 1}^{\infty} \frac{1}{{(36 \times \frac{π^{2}}{6} - 1)}^{1}} S_{2} = \sum_{k = 1}^{\infty} \frac{2}{{(36 \times \frac{π^{2}}{6} - 1)}^{2}}$

Step 2

If S₁ and S₂ is added , then the value will be ;

$S_{1} + S_{2} = \sum_{k = 1}^{\infty} \frac{1}{{(36 \times \frac{π^{2}}{6} - 1)}^{1}} + \sum_{k = 1}^{\infty} \frac{2}{{(36 \times \frac{π^{2}}{6} - 1)}^{2}} S_{1} + S_{2} = \sum_{k = 1}^{\infty} \frac{1}{(36 \times \frac{π^{2}}{6} - 1)} + \sum_{k = 1}^{\infty} \frac{2}{({(36 \times \frac{π^{2}}{6})}^{2} - 1^{2})} S_{1} + S_{2} = \sum_{k = 1}^{\infty} (\frac{1}{(6 π^{2} - 1)} + \frac{1}{({(6 π^{2})}^{2} - 1)})$

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