Not a century

$1^{2}+(1^{2}+2^{2})+(1^{2}+2^{2}+3^{2})+......\\=99*1^2+98*2^2+97*3^2+ . . . +1*99^2\\ \displaystyle= \sum_1^{99} (100-n)n^2\\\displaystyle= \sum_1^{99} (100n^2- n^3)\\=100*\dfrac{99*(99+1)(2*99+1)}{6}-\left (\dfrac{99*(99+1)} 2 \right )^2\\\color{#D61F06}{8332500}.$

The above expression can be represented by $\displaystyle\sum_{n=1}^{99}\sum_{k=1}^nk^2$ .

And $\color{#624F41}{\displaystyle\sum_{k=1}^nk^2=\dfrac{n(n+1)(2n+1)}6}$ .

Thus,

$\begin{aligned}\displaystyle\sum_{n=1}^{99}\color{#624F41}{\sum_{k=1}^nk^2}=&\ \displaystyle\sum_{n=1}^{99}\color{#624F41}{\dfrac{n(n+1)(2n+1)}6}\\=&\ \displaystyle\sum_{n=1}^{99}\dfrac{2n^3+3n^2+n}6\\=&\ \displaystyle\sum_{n=1}^{99}\left(\dfrac13n^3+\dfrac12n^2+\dfrac16n\right)\\=&\ \dfrac13\color{#D61F06}{\displaystyle\sum_{n=1}^{99}n^3}+\dfrac12\color{#20A900}{\displaystyle\sum_{n=1}^{99}n^2}+\dfrac16\color{#3D99F6}{\displaystyle\sum_{n=1}^{99}n}\\=&\ \dfrac13\color{#D61F06}{\left(\dfrac{99(100)}2\right)^2}+\dfrac12\color{#20A900}{\left(\dfrac{99(100)(199)}6\right)}+\dfrac16\color{#3D99F6}{\left(\dfrac{99(100)}2\right)}\\=&\ \dfrac13\color{#D61F06}{(24502500)}+\dfrac12\color{#20A900}{(328350)}+\dfrac16\color{#3D99F6}{(4950)}\\=&\ 8167500+164175+825=\boxed{8332500}\end{aligned}$

Note that $\sum_{k=1}^{n}k(k+1)(k+2)\ldots (k+m)=\frac{k(k+1)(k+2)\ldots (k+m)(k+m+1)}{m+2}$

Now $\displaystyle \sum_{k=1}^{99}\sum_{m=1}^{k}m^2 =\sum_{k=1}^{99} \frac{k(k+1)(2k+1)}{6} \\ \displaystyle=\frac{2}{6}\sum_{k=1}^{99} k(k+1)(k+2) - \frac{3}{6}\sum_{k=1}^{99} k(k+1)\\ \displaystyle=\frac{2}{6} \frac{99(100)(101)(102)}{4} - \frac{3}{6}\frac{99(100)(101)}{3}\\ \displaystyle=8332500$

You can use the following code: s = 0 k = 100 For i = 1 To 99 kk = k - i s = s + kk * i ^ 2 Next i

and get s=8332500