If sigma r =1 to n r=55 then find sigma r=1 to n r^3

according to your question you are asked sum of cubes of first 55 natural no.That is 1^3+2^3+3^3+4^3+...+55^3Sol:See there is a formular for sum of first n natural no cubes that is [n(n+1)/2]^2in this case n is 55 so sum=[55×56/2]^3=(55×28)^2=2371600I tried to answer you in simple way if correct please give approval.