Question

Thu gọn các biểu thức sau:

A=2^100-2^99+2^98-2^97+...+2-1

B=3+3^2+3^3+...+3^2017

C=1/2+1/2^2+1/2^3+...+!?2^1000

Giúp mình vs nhak. Đang cần gấp lắm.

Cảm ơn nhiều

Answer 1

\(A=2^{100}-2^{99}+2^{98}-2^{97}+2-1\\ 2A=2^{101}-2^{100}+2^{99}+...+2^2-2\\ 2A+A=\left(2^{101}-2^{100}+2^{99}+...+2^2-2\right)+\left(2^{100}-2^{99}+2^{98}-2^{97}+2-1\right)\\ 3A=2^{101}-1\\ A=\dfrac{2^{101}-1}{3}\)

\(B=3+3^2+3^3+...+3^{2017}\\ 3B=3^2+3^3+3^4+...+3^{2018}\\ 3B-B=\left(3^2+3^3+3^4+...+3^{2018}\right)-\left(3+3^2+3^3+...+3^{2017}\right)\\ 2B=3^{2018}-3\\ B=\dfrac{3^{2018}-3}{2}\)

\(C=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}\\ 2C=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{999}}\\ 2C-C=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{999}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}\right)\\ C=1-\dfrac{1}{2^{1000}}\)