Question

tính:

\(M=2-2^2+2^3-2^4+...+2^{19}-2^{20}\)

\(P=4^{100}-4^{98}+4^{96}-4^{94}+...+4^4-4^2\)

help mehMo Nguyễn Văn No choice teen Tạ Khánh Linh Dương Bá Gia Bảo Phạm Hoàng Hải Anh

Answer 1

\(M=2-2^2+2^3-2^4+...+2^{19}-2^{10}\\ 2.M=2^2-2^3+2^4-2^5+...+2^{20}-2^{21}\\ 2.M+M=2^{21}+2\\ \Rightarrow M=\frac{2^{21}+2}{3}\)

Answer 2

\(P=4^{100}-4^{98}+4^{96}-4^{94}+...+4^4-4^2\\ 4^2.P=4^{102}-4^{100}+4^{98}-4^{96}+...+4^6-4^4\\ 4^2.P+P=4^{102}-4^2\\ \Rightarrow P=\frac{4^{102}-4^2}{17}\)

Answer 3

M = 2 - 2^2 + 2^3 - 2^4 + ... + 2^19 - 2^20 2M = 2^2 - 2^3 + 2^4 - ... - 2^19 + 2^20 - 2^21 2M +M = -2^21 + 2 3M = -2^21 + 2 M = (-2^21 + 2) : 3

Answer 4

P = 4^100 - 4^98 + 4^96 - 4^94 + ... + 4^4 - 4^2

16P = 4^102 - 4^100 + 4^98 - 4^96 + 4^94 - ... - 4^8 + 4^6 - 4^4

16P + P = (4^102 - 4^100 + 4^98 - 4^96 + 4^94 - ... - 4^8 + 4^6 - 4^4) + (4^100 - 4^98 + 4^96 - 4^94 + ... + 4^4 - 4^2)

17P = 4^102 - 4^2

P = (4^102 - 4^2):17