If nth term of a series is 1/2 ×n(n+1) then find the sum of n terms of the series ?

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Answer 1 · 2018-08-07T18:12:50

$S_{n} = \frac{n}{6} (n + 1) (n + 2)$ $S_n=n/6(n+1)(n+2)$

Here,

$n^{t h} t e r m$ $n^(th) term$ of series is : $\frac{n}{2} (n + 1)$ $n/2(n+1)$

So, the sum of first terms of series is:

$S_n=1+(1+2)+(1+2+3)+...+n/2(n+1)$

$:.S_n=sum_(r=1)^n r/2(r+1)=1/2sum_(r=1)^n(r^2+r)$

$:.S_n=1/2{sum_(r=1)^nr^2+sum_(r=1)^nr}$

$:.S_n=1/2{n/6(n+1)(2n+1)+n/2(n+1)}$

$:.S_n=1/2*n/2(n+1){(2n+1)/3+1}$

$:.S_n=n/4(n+1){(2n+1+3)/3}$

$:.S_n=n/12(n+1){2n+4}$

$:.S_n=n/12(n+1)2(n+2)$

$:.S_n=n/6(n+1)(n+2)$