Question

What is the nth term of:
1^2 + (1^2+2^2) + (1^2+2^2+3^2)+.....

Answer 1

\[an = \frac{ n(n+1)(2n+1) }{ 6 }\]

Answer 2

how??

Answer 3

Do you see how the nth term is something like
\[1^2+2^2+\cdots+(n-1)^2+n^2?\]
What's the sum of all square numbers from 1 to n? i.e.,
\[\sum_{k=1}^n k^2?\]