If Sn-132-1-142-2-152-3-upto n terms and S- a72- 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+ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Method of Difference

If Sn=132+1+1...

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

Open in App

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 $n,$
$\frac{1}{n + 2}, \frac{1}{n + 3}, \frac{1}{n + 4} ≅ 0$
$S_{\infty} = \frac{1}{3} [\frac{1}{2} + \frac{1}{3} + \frac{1}{4}] = \frac{13}{36} ∴ a = 26$

Suggest Corrections

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

and

S_{\infty} = \frac{a}{72},

then

a

equals

Q. If

S = \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . . . . . . . . . . .

\infty

then find

\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \frac{1}{7^{2}} + . . . . . . . . . . . . .

\infty

in terms of S.

Q. If

x = \frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . . . .,

y = \frac{1}{1^{2}} + \frac{3}{2^{2}} + \frac{1}{3^{2}} + \frac{3}{4^{2}} + . .

and

z = \frac{1}{1^{2}} - \frac{1}{2^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + . . . .,

then

Q. If

\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . . \infty = \frac{π^{2}}{6}

then,

\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . . . . \infty

is equal to

Q. The value of the sum

\frac{1}{3^{2} + 1} + \frac{1}{4^{2} + 2} + \frac{1}{5^{2} + 3} + \frac{1}{6^{2} + 4} + . . . . . \infty

is equals to -

Join BYJU'S Learning Program

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