If S-n-2-3n2-1n2-13- then 16S is

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) ⇒ nbs ...

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Byju's Answer

Standard XII

Mathematics

Vn Method

If S=∑n=2∞3n2...

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

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Similar questions

Q. If

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

then

16 S

\frac{9}{- 13} + \frac{4}{- 16} = \frac{4}{- 16} +

\frac{9}{- 13}

.
If true then enter

1

and if false then enter

0

n L t \to \infty [\frac{1}{n} + \frac{n^{2}}{(n + 1)^{3}} + \frac{n^{2}}{(n + 2)^{3}} + \dots . + \frac{1}{8 n}] =

$lim n \to \infty (\frac{1}{n} + \frac{n^{2}}{(n + 1)^{3}} + \frac{n^{2}}{(n + 2)^{3}} + . . . + \frac{1}{8 n})$ is equal to

Q. If

S = {tan}^{- 1} (\frac{1}{n^{2} + n + 1}) + {tan}^{- 1} (\frac{1}{n^{2} + 3 n + 3}) + \dots + {tan}^{- 1} (\frac{1}{1 + (n + 19) (n + 20)})

then

tan S

is equal to

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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