Ytukyg

Find the close expression for :
$$sum_{k=1}^nleft(sum_{l=1}^k l^2right)$$

I have very less knowledge about nested sigma notations. Any help or hints would be greatly appreciated.

To evaluate this sum by hand efficiently, we can use something calling rising factorial.

For any $n > 0$, the rising factorial is a polynomial in $x$ defined by:

$$x^{(n)} = prodlimits_{k=0}^{n-1}(x+k)$$
One useful property of $x^{(n)}$ is it can be expressed as a difference of successive terms of $x^{(n+1)}$

$$begin{align}x^{(n)}
&= x(x+1)cdots(x+n-1)\ &= x(x+1)cdots(x+n-1)frac{(x+n)-(x-1)}{n+1}\
&= frac{1}{n+1}left(x^{(n+1)} - (x-1)^{(n+1)}right)end{align}
$$
This means summation over $x^{(n)}$ is a telescoping one.
$$begin{align}sum_{x=1}^y (x)_n &= frac{1}{n+1}sum_{x=1}^yleft( x^{(n+1)} - (x-1)^{(n+1)}right)
= frac{1}{n+1}left( y^{(n+1)} - 0^{(n+1)}right)\
&= frac{1}{n+1} y^{(n+1)}end{align}$$

For the sum at hand, notice $ell^2 = ell(ell+1) - ell = ell^{(2)} - ell^{(1)}$, we have

$$begin{align}sum_{k=1}^n sum_{ell=1}^k ell^2
&= sum_{k=1}^n sum_{ell=1}^kleft( ell^{(2)} - ell^{(1)}right)
= sum_{k=1}^n left(frac13 k^{(3)} - frac12 k^{(2)}right)\
&= frac{1}{3cdot 4}n^{(4)} - frac{1}{2cdot 3} n^{(3)}
= frac{n(n+1)(n+2)(n+3 - 2)}{12}\
&= frac{n(n+1)^2(n+2)}{12}end{align}$$

A brute force way is as follows. The inner sum can be written as
$$
sum_{l=1}^k l^2=frac{k(k+1)(2k+1)}{6}tag{1}
$$
To sum this over $k$ expand the LHS of $(1)$ and use the formulae
$$
sum_{j=1}^nj=frac{n(n+1)}{2}quad sum_{j=1}^nj^2=frac{n^2(n+1)^2}{4}.
$$

Hint:

A sum of $k^{th}$ powers can be expressed as a $k+1^{th}$ degree polynomial, quartic in your case. Furthermore, $P(0)=0$ because a void summation is zero, and

$$S(n)=n(an^3+bn^2+cn+d)$$ or

$$Q(n):=frac{P(n)}n=an^3+bn^2+cn+d$$ can be identified as the Lagrangian interpolation polynomial through the points $left(i,dfrac{S(i)}iright)$ for $i=1,2,3,4$.