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Mathematics Question on Sequence and series

The sum of the first n terms 121+12+221+2+12+22+321+2+3+ \frac {1^2}{1} +\frac {1^2+2^2}{1+2}+ \frac {1^2 +2^2+3^2}{1+2+3}+ …\dots is

A

n2−2n3\frac {n^2-2n}{3}

B

2n2+n3\frac {2n^2+n}{3}

C

n(n+2)3\frac {n(n+2)}{3}

D

2n2−n3\frac {2n^2-n}{3}

Answer

n(n+2)3\frac {n(n+2)}{3}

Explanation

Solution

Given series, 121+12+221+2+12+22+321+2+3+...\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+... The nn th terms of the series is Tn=Σn2Σn=n(n+1)(2n+1)6n(n+1)2=(2n+1)3T_{n}= \frac{\Sigma n^{2}}{\Sigma n}=\frac{n(n+1)(2 n+1)}{\frac{6 n(n+1)}{2}}=\frac{(2 n+1)}{3} Now, Sn=13(2Σn+Σ1)S_{n} =\frac{1}{3}( 2\Sigma n+\Sigma 1) =\frac{1}{3}\left\\{2 \cdot \frac{n(n+1)}{2}+n\right\\}=\frac{n}{3}(n+1+1) =n(n+2)3=\frac{n(n+2)}{3}