Question

find sum to n terms 1^2+(1^2+2^2)+(1^2+2^2+3^2)......

Answer 1

terms are:1,5,9,13....
a=1. d=4.

Answer 2

Here in thus series the last term should be
(1^2 + 2^2 + 3^2...........n^2)

this last term is equal to
n(n+1)(2n+1)/6. [the sum of square of first n terms]

so open the bracket we have
(2n^3 + 3n^2 + n)/6 ........(1)

now, use the formula for each to get the sum of n terms

2n^3 = 2[n(n+1)/6]^2. (sum of cube of first n tetms)

3n^2 = 3[n(n+1)(2n+1)/6] (sum of square of first n terms)

n = n(n+1)/2. sumof first n terms

now put these value in equation (1)

[2n^2(n+1)^2/4 + 3n(n+1)(2n+1)/6 + n(n+1)/2]/6

[n(n+1)/2 {2n(n+1)/2 + 3(2n+1)/3 + 1}]/6

n(n+1)/12 {6n(n+1) + 6(2n+1) + 6}/6
n(n+1)/12{6n^2 + 6n + 12n + 6 + 6}

n(n+1)/12 {[6n^2 + 18n 12]/6}

n(n+1)/12{6[n^2 + 3n + 2]/6}

n(n+1)(n^2 + 3n + 2)/12

n(n+1)(n+1)(n+2)/12

n(n+1)^2(n+2)/12

it is the answer of that question