What is the value of the sum: #(1^2)+(1^2+2^2)+(1^2+2^2+3^2)+............#upto n terms? Thank you!
1 Answer
Nov 24, 2016
Explanation:
We will use the following known sums (each of which can be proven via induction):
#sum_(i=1)^ni = (n(n+1))/2# #sum_(i=1)^ni^2=(n(n+1)(2n+1))/6# #sum_(i=1)^n(i^3)=(n^2(n+1)^2)/4#
With those:
#= sum_(i=1)^n(1^2+2^2+...+i^2)#
#=sum_(i=1)^nsum_(j=1)^ij^2#
#= sum_(i=1)^n(i(i+1)(2i+1))/6#
#=sum_(i=1)^n(2i^3+3i^2+i)/6#
#=sum_(i=1)^n1/3i^3+sum_(j=1)^n1/2j^2+sum_(k=1)^n1/6k#
#=1/3sum_(i=1)^ni^3+1/2sum_(j=1)^nj^2+1/6sum_(k=1)^nk#
#=1/3((n^2(n+1)^2)/4)+1/2((n(n+1)(2n+1))/6) + 1/6((n(n+1))/2)#
#=(n^2(n+1)^2)/12+(n(n+1)(2n+1))/12+(n(n+1))/12#
#=(n(n+1)[n(n+1)+(2n+1)+1])/12#
#=(n(n+1)(n^2+3n+2))/12#
#=(n(n+1)^2(n+2))/12#
