Squares

A difference of squares $x^2-y^2$ can be factored as $(x-y)(x+y)$ .

$\Rightarrow 1^2-2^2+3^2-4^2+\ldots+457^2-458^2 =$ $(1-2)(1+2)+(3-4)(4-3)+\ldots+(457-458)(457+458)$

The $x+y$ terms follow an arithmetic progression that can be stated as $\sum\limits_{n=1}^{229} 3+4(n-1)$ = $229 \cdot \frac{3+915}{2} = 105111$ .

But don't forget! Since all of the $x-y$ terms are $-1$ , the final answer is negative.

$\Rightarrow \boxed{-105111 }$

by taking each a^2-b^2 as one term you get an arithmetic progression of -3+-7+-11........+-915...... each terms differ by d=-4...... total term is 229..... therefore use AP fromula ...... you get answer......

Let f(x) = x^2 - (x+1)^2

f(x) = (x-x-1)(x+x+1)

f(x) = -(2x+1)

The question then can be written as:

f(1) + f(3) + ...... + f(457)

= -3-7-11-……-915

= -(3+7+11+…..+915)

= -(229/2 (3+915)

= -229/2 (918)

= -105111