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

Question

2201 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-07T18:12:50

#S_n=n/6(n+1)(n+2)#

Here,

#n^(th) term# of series is :#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)#