An algebra problem by Aziz Alasha

The $i^{th}$ term of the series is $1^2+2^2+3^2+...+i^2=\dfrac{2i^3+3i^2+i}{6}$ . So $S_n=$ sum of first $n$ terms of the series is $\dfrac{n(n+1)^2(n+2)}{12}$ . Therefore the sum of the first $1000$ terms is $\boxed {83667083500}$

the sum will be "a sigma of sigmas"

the nth term will beUn = ∑n² = n(n+1)(2n+1)/6

S(n) = ∑Un = ∑n(n+1)(2n+1)/6 = n(n+1)²(n+2)/12

S(1000) = 83666083500

$\begin{aligned} S & = 1^2 + (1^2+2^2) + (1^2+2^2+3^2) + \cdots + (1^2+2^2 + \cdots + 1000^2) \\ & = 1000\times 1^2 + 999 \times 2^2 + 998 \times 3^2 + \cdots + 1\times 1000^2 \\ & = \sum_{n=1}^{1000} (1001-n)n^2 \\ & = 1001 \sum_{n=1}^{1000} n^2 - \sum_{n=1}^{1000} n^3 \\ & = 1001 \times \frac {1000(1001)(2001)}6 - \left(\frac {1000(1001)}2\right)^2 \\ & = \boxed{83667083500} \end{aligned}$