If S - n -2-3 n 2-1- n 2-13 then 16 S i

Given : nbsp;S=∑_n=2^∞3n^2+1(n^2 1)^3 We know that, (n^2 1)^3=(n 1)^3(n+1)^3 (n+1)^3 (n 1)^3=6n^2+2 Now, ⇒ S=∑_n=2^∞12((n+1)^3 (n 1)^3(n+1)^3(n 1)^3) ⇒ n ...

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Standard XII

Mathematics

Vn Method

If S =∑ n =2∞...

Question

If $S = \infty \sum n = 2 \frac{3 n^{2} + 1}{(n^{2} - 1)^{3}}$ then $16 S$ is

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Solution

Given : $S = \infty \sum n = 2 \frac{3 n^{2} + 1}{(n^{2} - 1)^{3}}$

We know that,
$(n^{2} - 1)^{3} = (n - 1)^{3} (n + 1)^{3} (n + 1)^{3} - (n - 1)^{3} = 6 n^{2} + 2$

Now,
$\Rightarrow S = \infty \sum n = 2 \frac{1}{2} (\frac{(n + 1)^{3} - (n - 1)^{3}}{(n + 1)^{3} (n - 1)^{3}}) \Rightarrow S = \frac{1}{2} \infty \sum n = 2 (\frac{1}{(n - 1)^{3}} - \frac{1}{(n + 1)^{3}}) \Rightarrow S = \frac{1}{2} [(\frac{1}{1^{3}} - \frac{1}{3^{3}}) + (\frac{1}{2^{3}} - \frac{1}{4^{3}}) + (\frac{1}{3^{3}} - \frac{1}{5^{3}}) + \dots] \Rightarrow S = \frac{1}{2} (\frac{1}{1^{3}} + \frac{1}{2^{3}}) \Rightarrow S = \frac{9}{16} ∴ 16 S = 9$

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If S=∞∑n=23n2+1(n2−1)3 then 16S is

Given : S=∞∑n=23n2+1(n2−1)3 We know that, (n2−1)3=(n−1)3(n+1)3(n+1)3−(n−1)3=6n2+2 Now, ⇒S=∞∑n=212((n+1)3−(n−1)3(n+1)3(n−1)3)⇒S=12∞∑n=2(1(n−1)3−1(n+1)3)⇒S=12[(113−133)+(123−143)+(133−153)+…]⇒S=12(113+123)⇒S=916∴16S=9

If $S = \infty \sum n = 2 \frac{3 n^{2} + 1}{(n^{2} - 1)^{3}}$ then $16 S$ is