Найдите сумму 1² - 2² + 3² - 4² +...+ 99² - 100²

$A = \sum\limits_{i = 1}^{100} (-1)^{i+1}i^2 = 1^2 - 2^2 + 3^2 - 4^2 + ... + 99^2 - 100^2\\\\ \left[ \ \ 1 + 3 + ... + (2n - 1) = n^2 \ \ \right]\\\\ 1 - (1 + 3) + (1 + 3 + 5) - (1 + 3 + 5 + 7) + ... + (1 + ... + 197) - \\\\ -(1 + ... + 197 + 199) = (1 - 1) - 3 + (1+3 + 5 - 1 - 3 - 5) - 7 + ...\\\\ ...+(1+...+197-1-...-197)-199 = -(3 + 7 +...+199) =\\\\$

$= -\sum\limits_{j = 1}^{50}(4j-1) = -4\sum\limits_{j = 1}^{50}(j) + 50 = \left[ \ 1 + 2 + ... + n = \frac{n(n+1)}{2}\ \right] = \\\\ = -4*\frac{50*51}{2} + 50 = -5100 + 50 = \boxed{-5050}$

Немного отличное решение:

$A = \sum\limits_{i = 1}^{100} (-1)^{i+1}i^2 = 1^2 - 2^2 + 3^2 - 4^2 + ... + 99^2 - 100^2 = \\\\ = -((2^2 - 1^2) + (4^2 - 3^2) + ... + (100^2 - 99^2)) = \\\\ \left[ \ (n+1)^2 - n^2 = 2n+1 \ \right]\\\\ = -(3 + 7 + ... + 199) = ...= \boxed{-5050}$

Более того:

$3 = 1 + 2, \ 7 = 3 + 4, \ 11 = 5 + 6, ..., \ 199 = 99 + 100\\\\ \left[ \ 4n - 1 = (2n - 1) + 2n\ \right]\\\\ -\sum\limits_{j = 1}^{50}(4j-1) = -\sum\limits_{j = 1}^{100}j = -(1 + 2 + ... + 100) = \\\\ = -((1 + 100) + (2 + 99) + ... + (50 + 51)) = -50*101 = -5050$