Câu hỏi của Thân Trung Hiếu

Question

Thu gọn K= 2^1-2^2+2^3-2^4+...+2^99-2^100

Mashiro Shiina · Accepted Answer

$K=2^1-2^2+2^3-2^4+...+2^{99}-2^{100}$

$2K=2\left(2^1-2^2+2^3-2^4+...+2^{99}-2^{100}\right)$

$2K=2^2-2^3+2^4-2^5+....+2^{100}-2^{101}$

$2K+K=\left(2^2-2^3+2^4-2^5+.....+2^{100}-2^{101}\right)+\left(2^1-2^2+2^3-2^4+.....+2^{99}-2^{100}\right)$$3K=2-2^{101}$

$K=\dfrac{2-2^{101}}{3}$Câu trả lời: $K=2^1-2^2+2^3-2^4+...+2^{99}-2^{100}$

$2K=2\left(2^1-2^2+2^3-2^4+...+2^{99}-2^{100}\right)$

$2K=2^2-2^3+2^4-2^5+....+2^{100}-2^{101}$

$2K+K=\left(2^2-2^3+2^4-2^5+.....+2^{100}-2^{101}\right)+\left(2^1-2^2+2^3-2^4+.....+2^{99}-2^{100}\right)$$3K=2-2^{101}$

$K=\dfrac{2-2^{101}}{3}$

Đại số lớp 6