The value of 12 - r -1-12 r 2-1-64 r 6-48 r 4-12 r 2-1 is

nbsp;12∑^∞_r=112r^2+164r^6 48r^4+12r^2 1 =6∑^∞_r=124r^2+2/(4r^2 1)^3 =6∑^∞_r=1T_r nbsp; T_r=24r^2+2(2r 1)^3(2r+1)^3 =(2r+1)^3 (2r 1)^3(2r 1)^3(2r+1)^3 =1( ...

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Standard XII

Mathematics

Summation by Sigma Method

The value of ...

Question

The value of $12 \infty \sum r = 1 \frac{12 r^{2} + 1}{64 r^{6} - 48 r^{4} + 12 r^{2} - 1}$ is

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Solution

$12 \infty \sum r = 1 \frac{12 r^{2} + 1}{64 r^{6} - 48 r^{4} + 12 r^{2} - 1}$ $= 6 \infty \sum r = 1 \frac{24 r^{2} + 2}{(4 r^{2} - 1)^{3}} = 6 \infty \sum r = 1 T_{r}$
$T_{r} = \frac{24 r^{2} + 2}{(2 r - 1)^{3} (2 r + 1)^{3}} = \frac{(2 r + 1)^{3} - (2 r - 1)^{3}}{(2 r - 1)^{3} (2 r + 1)^{3}} = \frac{1}{(2 r - 1)^{3}} - \frac{1}{(2 r + 1)^{3}} \Rightarrow 6 \infty \sum r = 1 T_{r} = 6$

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The value of 12∞∑r=112r2+164r6−48r4+12r2−1 is

12∞∑r=112r2+164r6−48r4+12r2−1 =6∞∑r=124r2+2(4r2−1)3=6∞∑r=1Tr Tr=24r2+2(2r−1)3(2r+1)3=(2r+1)3−(2r−1)3(2r−1)3(2r+1)3=1(2r−1)3−1(2r+1)3⇒6∞∑r=1Tr=6

The value of $12 \infty \sum r = 1 \frac{12 r^{2} + 1}{64 r^{6} - 48 r^{4} + 12 r^{2} - 1}$ is