If S n -1-32-1-1-42-2-1-52-3- upto n terms and S - a -72- then a equals

T_r = 1r + (r+2)^2 = 1(r+4)(r+1) =13 [ 1r+1 1r+4 ] S_n=∑_r=1^nT_r = 13 ( 12 15 ) +13 ( 13 16 ) +13 ( 14 17 ) +13 ( 15 18 ) +13(16 19) +13(⋮ ⋮) +13 [ 1n+ ...

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Byju's Answer

Standard XI

Mathematics

Continuity of a Function

If S n =1/32+...

Question

If $S_{n} = \frac{1}{3^{2} + 1} + \frac{1}{4^{2} + 2} + \frac{1}{5^{2} + 3} + \dots \dots upto n terms$ and $S_{\infty} = \frac{a}{72},$ then $a$ equals

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Solution

$T_{r} = \frac{1}{r + (r + 2)^{2}} = \frac{1}{(r + 4) (r + 1)}$
$= \frac{1}{3} [\frac{1}{r + 1} - \frac{1}{r + 4}]$
$S_{n} = n \sum r = 1 T_{r} = \frac{1}{3} (\frac{1}{2} - \frac{1}{5})$
$+ \frac{1}{3} (\frac{1}{3} - \frac{1}{6})$
$+ \frac{1}{3} (\frac{1}{4} - \frac{1}{7})$
$+ \frac{1}{3} (\frac{1}{5} - \frac{1}{8}) + \frac{1}{3} (\frac{1}{6} - \frac{1}{9}) + \frac{1}{3} (⋮ ⋮) + \frac{1}{3} [\frac{1}{n + 1} - \frac{1}{n + 4}] S_{n} = \frac{1}{3} [\frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{n + 2} - \frac{1}{n + 3} - \frac{1}{n + 4}]$
For very large value of $n,$
$\frac{1}{n + 2}, \frac{1}{n + 3}, \frac{1}{n + 4} ≅ 0$
$S_{\infty} = \frac{1}{3} [\frac{1}{2} + \frac{1}{3} + \frac{1}{4}] = \frac{13}{36} ∴ a = 26$

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Similar questions

Q. If

S_{n} = \frac{1}{3^{2} + 1} + \frac{1}{4^{2} + 2} + \frac{1}{5^{2} + 3} + \dots \dots upto n terms

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Show by the Principle of Mathematical induction that the sum $S_{n}$ of the n terms of the series $1^{2} + 2 \times 2^{2} + 3^{2} + 2 \times 4^{2} + 5^{2} + 2 \times 6^{2} + 7^{2} + . . . . .$ is given by $S_{n} = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{n (n + 1)^{2}}{2}, if n is even \frac{n^{2} (n + 1)}{2}, if n is odd \end{matrix}$

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If Sn=132+1+142+2+152+3+⋯⋯upto n terms and S∞=a72, then a equals

Tr=1r+(r+2)2=1(r+4)(r+1) =13[1r+1−1r+4] Sn=n∑r=1Tr=13(12−15) +13(13−16) +13(14−17) +13(15−18)+13(16−19)+13(⋮ ⋮)+13[1n+1−1n+4]Sn=13[12+13+14−1n+2−1n+3−1n+4] For very large value of n, 1n+2,1n+3,1n+4≅0 S∞=13[12+13+14]=1336∴a=26

If $S_{n} = \frac{1}{3^{2} + 1} + \frac{1}{4^{2} + 2} + \frac{1}{5^{2} + 3} + \dots \dots upto n terms$ and $S_{\infty} = \frac{a}{72},$ then $a$ equals