Question

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

Answer 1

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

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

Немного отличное решение:

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

Более того:

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