Sum of the series $1^2-2^2+3^2-4^2+......2001^2-2002^2+2003^2$

Question

Sum of the series $1^2-2^2+3^2-4^2+......2001^2-2002^2+2003^2$

How to solve this series?

$(1^2+3^2+5^2+ \cdots +2003^2)-(2^2+4^2+6^2+ \cdots +2002^2)$

Now what to do next?

Answer 1

AP only: $$(1^2-0^2) + (3^2-2^2)+\ldots + (2003^2-2002^2)$$ $$=1\cdot 1 + 1\cdot 5 + \ldots + 1\cdot 4005$$ $$=1+5+9+\ldots + 4005 $$ $$= \ldots$$

Answer 2

$$\sum_{n=0}^N (n+1) r^n = \dfrac{d}{dr} \sum_{n=0}^N r^{n+1}$$

$$\sum_{n=0}^N (n+1)^2 r^n = \dfrac{d}{dr} \sum_{n=0}^N (n+1) r^{n+1} = \dfrac{d}{dr} \left(r \sum_{n=0}^N (n+1) r^n \right)$$