Find the sum of first 10 terms of following series S -31-1-513-23-12-22-713-23-33-12-22-32-A- 440B- 450C- 660D- 220

The correct option is C 660First we shall find the r^th term of the given series T_r=(2r+1)[r(r+1)/2]^2/r(r+1)(2r+1)/6=3r(r+1)/2 T_r=3/2[r^2+r] S_n=∑ T_r=3/ ...

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Question

Find the sum of first $10$ terms of following series
$S = \frac{3 (1)}{1} + \frac{5 (1^{3} + 2^{3})}{1^{2} + 2^{2}} + \frac{7 (1^{3} + 2^{3} + 3^{3})}{1^{2} + 2^{2} + 3^{2}} + . . . .$

440

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450

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660

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220

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Solution

The correct option is C $660$
First we shall find the $r^{th}$ term of the given series

$T_{r} = \frac{(2 r + 1) {[\frac{r (r + 1)}{2}]}^{2}}{\frac{r (r + 1) (2 r + 1)}{6}} = \frac{3 r (r + 1)}{2}$

$T_{r} = \frac{3}{2} [r^{2} + r]$

$S_{n} = \sum T_{r} = \frac{3}{2} [\frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}]$

$S_{10} = \frac{3}{2} [\frac{10 \times 11 \times 21}{6} + \frac{10 \times 11}{2}]$

$S_{10} = \frac{3}{2} [385 + 55] = 660$

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Q. Find the sum of first

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Q. Let

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Find the sum of first 10 terms of following series S=3(1)1+5(13+23)12+22+7(13+23+33)12+22+32+....

The correct option is C 660First we shall find the rth term of the given series Tr=(2r+1)[r(r+1)2]2r(r+1)(2r+1)6=3r(r+1)2 Tr=32[r2+r] Sn=∑Tr=32[n(n+1)(2n+1)6+n(n+1)2] S10=32[10×11×216+10×112] S10=32[385+55]=660

Find the sum of first $10$ terms of following series
$S = \frac{3 (1)}{1} + \frac{5 (1^{3} + 2^{3})}{1^{2} + 2^{2}} + \frac{7 (1^{3} + 2^{3} + 3^{3})}{1^{2} + 2^{2} + 3^{2}} + . . . .$