S - 1-frac12-lfrac16-frac112-frac120- infty 1 If - S - 1-2 S - 2-frac a 6- - text then - a - V

S_1=1/2+1/6+1/12+1/20+...∞ =∑_n=1^∞1/n(n+1) =∑_n=1^∞1/n 1/n+1 =∑_n=1^∞(1/1 1/2)+(1/2 1/3)+(1/3 1/4)+...+(1/n 1/n+1) =lim_n →∞(1 1/n+1) S_1=1 lim_n →∞1/n+1= ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

S - 1=frac12+...

Question

$S_{1} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + . . . \infty S_{2} = \frac{1}{3} + \frac{1}{5^{2}} + \frac{1}{3^{3}} + \frac{1}{5^{4}} + \frac{1}{3^{5}} + \frac{1}{5^{6}} + . . . \infty If S_{1} - 2 S_{2} = \frac{a}{6}, then a =$

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Solution

$S_{1} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + . . . \infty = \infty \sum n = 1 \frac{1}{n (n + 1)} = \infty \sum n = 1 \frac{1}{n} - \frac{1}{n + 1} = \infty \sum n = 1 (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + . . . + (\frac{1}{n} - \frac{1}{n + 1}) = lim n \to \infty (1 - \frac{1}{n + 1}) S_{1} = 1 - lim n \to \infty \frac{1}{n + 1} = 1 S_{2} = \frac{1}{3} + \frac{1}{5^{2}} + \frac{1}{3^{3}} + \frac{1}{5^{4}} + \frac{1}{3^{5}} + \frac{1}{5^{6}} + . . . \infty = (\frac{1}{3} + \frac{1}{3^{3}} + \frac{1}{3^{5}} + . . . \infty) + (\frac{1}{5^{2}} + \frac{1}{5^{4}} + \frac{1}{5^{6}} + . . . \infty) = \frac{\frac{1}{3}}{1 - \frac{1}{3^{2}}} + \frac{\frac{1}{5^{2}}}{1 - \frac{1}{5^{2}}} = \frac{3}{8} + \frac{1}{24} = \frac{5}{12} S_{1} - 2 S_{2} = 1 - \frac{5}{6} = \frac{1}{6} \Rightarrow a = 1$

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S1=12+16+112+120+...∞S2=13+152+133+154+135+156+...∞If S1−2S2=a6, then a=

S1=12+16+112+120+...∞=∞∑n=11n(n+1)=∞∑n=11n−1n+1=∞∑n=1(11−12)+(12−13)+(13−14)+...+(1n−1n+1)=limn→∞(1−1n+1)S1=1−limn→∞1n+1=1S2=13+152+133+154+135+156+...∞=(13+133+135+...∞)+(152+154+156+...∞)=131−132+1521−152=38+124=512S1−2S2=1−56=16⇒a=1

$S_{1} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + . . . \infty S_{2} = \frac{1}{3} + \frac{1}{5^{2}} + \frac{1}{3^{3}} + \frac{1}{5^{4}} + \frac{1}{3^{5}} + \frac{1}{5^{6}} + . . . \infty If S_{1} - 2 S_{2} = \frac{a}{6}, then a =$