Question

Find the sum to n terms of the series
1² +(1²+2²)+(1²+2²+3²) +....​

Answer 1

Answer:

12+(12+22)+(12+22+32)+⋯+(12+22+32+⋯+n2)

=∑k=1n(12+⋯+k2)

=∑k=1n16k(k+1)(2k+1)

=16∑k=1n(2k3+3k2+k)

=13∑k=1nk3+12∑k=1nk2+16∑k=1nk

=112(n(n+1))2+112n(n+1)(2n+1)+112n(n+1)

=n(n+1)12(n(n+1)+(2n+1)+1)

=112n(n+1)2(n+2). ■

hope it's help you ✌️

Answer 2

Step-by-step explanation:

See the attachment

Hope , it helps you