The sum of the series 1 2 .2 + 2 2 .3 + 3 2 .4 + ........ to n terms is Options: A . &nbsp n 3 ( n + 1 ) 3 ( 2 n + 1 ) 24 B . &nbsp n ( n + 1 ) ( 3 n 2 + 7 n + 2 ) 12 C . &nbsp n ( n + 1 ) 6 [n(n+1) + (2n + 1)] D . &nbsp n ( n + 1 ) 12 [6n(n+1) + 2(2n + 1)]

Answer: Option B : B T π = n 2 (n + 1) = n 3 + n 2 S π = ∑ T π = ∑ n 3 + ∑ n 2 = [ n ( n + 1 ) 2 ] 2 + n ( n + 1 ) ( 2 n + 1 ) 6 = n ( n + 1 ) 2 [ n ( n + 1 ) 2 + 2 n + 1 3 ] = n ( n + 1 ) ( 3 n 2 + 7 n + 2 ) 12

The sum of the series 12.2 + 22.3 + 32.4 + ........ to n ter

Discussion Forum : Sequences And Series

Question -

The sum of the series $1^{2}$ .2 + $2^{2}$ .3 + $3^{2}$ .4 + ........ to n terms is

Options:

A .

\frac{n^{3} {(n + 1)}^{3} (2 n + 1)}{24}

B .

\frac{n (n + 1) (3 n^{2} + 7 n + 2)}{12}

C .

\frac{n (n + 1)}{6}

[n(n+1) + (2n + 1)]

D .

\frac{n (n + 1)}{12}

[6n(n+1) + 2(2n + 1)]

Answer: Option B
:
B

$T_{π}$ = $n^{2}$ (n + 1) = $n^{3}$ + $n^{2}$

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

= $\frac{n (n + 1)}{2}$ [ $\frac{n (n + 1)}{2}$ + $\frac{2 n + 1}{3}$ ] = $\frac{n (n + 1) (3 n^{2} + 7 n + 2)}{12}$

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Discussion Forum : Sequences And Series

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