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

Answer:

#(1^2)+(1^2+2^2)+...+(1^2+2^2+...+n^2)=(n(n+1)^2(n+2))/12#

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:

#(1^2)+(1^2+2^2)+...+(1^2+2^2+...+n^2)#

#= 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#