The sum of 10 terms of the series 312 - 22 - 522 - 32-732 - 42- is

T_n = 2n+1n^2 (n+1)^2 = (n+1)^2 n^2(n+1)^2n^2 S_n =∑^10_n=1 T_n = ∑^10_n=1(1n^2 1(n+1)^2) = (1 12^2)+ (12^2 13^2) + ⋯ + (110^2 111^2) = 1 1(11)^2 = ...

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Byju's Answer

Standard XII

Mathematics

Distance Formula Using Complex Numbers

The sum of 10...

Question

The sum of $10$ terms of the series $\frac{3}{1^{2} \times 2^{2}} + \frac{5}{2^{2} \times 3^{2}} + \frac{7}{3^{2} \times 4^{2}} + \dots$ is

\frac{143}{144}

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\frac{99}{100}

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\frac{120}{121}

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Solution

The correct option is C $\frac{120}{121}$
$T_{n} = \frac{2 n + 1}{n^{2} (n + 1)^{2}} = \frac{(n + 1)^{2} - n^{2}}{(n + 1)^{2} n^{2}}$
$S_{n} = 10 \sum n = 1 T_{n} = 10 \sum n = 1 (\frac{1}{n^{2}} - \frac{1}{(n + 1)^{2}})$
$= (1 - \frac{1}{2^{2}}) + (\frac{1}{2^{2}} - \frac{1}{3^{2}}) + \dots + (\frac{1}{10^{2}} - \frac{1}{11^{2}})$
$= 1 - \frac{1}{(11)^{2}} = \frac{120}{121}$

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The sum of 10 terms of the series 312×22+522×32+732×42+⋯ is

The correct option is C 120121Tn=2n+1n2(n+1)2=(n+1)2−n2(n+1)2n2 Sn=10∑n=1Tn=10∑n=1(1n2−1(n+1)2) =(1−122)+(122−132)+⋯+(1102−1112) =1−1(11)2=120121

The sum of $10$ terms of the series $\frac{3}{1^{2} \times 2^{2}} + \frac{5}{2^{2} \times 3^{2}} + \frac{7}{3^{2} \times 4^{2}} + \dots$ is