Q.Find $\displaystyle\sum_{k=1}^{15} k^2$.

Three closed-form sums recur constantly, both within this chapter and later in the syllabus: $\sum_{k=1}^n k=\dfrac{n(n+1)}{2}$ (the natural numbers viewed as an A.P. with $a=d=1$), $\sum_{k=1}^n k^2=\dfrac{n(n+1)(2n+1)}{6}$, and $\sum_{k=1}^n k^3=\left[\dfrac{n(n+1)}{2}\right]^2$. The last two are proved by telescoping: summing the identity $(k+1)^3-k^3=3k^2+3k+1$ (respectively $(k+1)^4-k^4=4k^3+6k^2+4k+1$) from $k=1$ to $n$ collapses the left side to a single difference of end-values, from wh …