Question

\(2017+\dfrac{2017}{2}+\dfrac{2017}{2^2}+\dfrac{2017}{2^3}+...+\dfrac{2017}{2^{2017}}\)

Answer 1

C1:

Gọi \(S=2017+\dfrac{2017}{2}+\dfrac{2017}{2^2}+\dfrac{2017}{2^3}+...+\dfrac{2017}{2^{2017}}\)

\(S=2017+\dfrac{2017}{2}+\dfrac{2017}{2^2}+\dfrac{2017}{2^3}+...+\dfrac{2017}{2^{2017}}\\ 2S=4034+2017+\dfrac{2017}{2^2}+...+\dfrac{2017}{2^{2016}}\\ 2S-S=\left(4034+2017+\dfrac{2017}{2^2}+...+\dfrac{2017}{2^{2016}}\right)-\left(2017+\dfrac{2017}{2}+\dfrac{2017}{2^2}+\dfrac{2017}{2^3}+...+\dfrac{2017}{2^{2017}}\right)\\ S=4034-\dfrac{2017}{2^{2017}}\)

(Khuyên dùng)

Answer 2

C2:

\(2017+\dfrac{2017}{2}+\dfrac{2017}{2^2}+\dfrac{2017}{2^3}+...+\dfrac{2017}{2^{2017}}\\ =2017\cdot\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2017}}\right)\)

Gọi \(S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2017}}\)

\(S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2017}}\\ 2S=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}\\ 2S-S=\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2017}}\right)\\ S=2-\dfrac{1}{2^{2017}}\)

\(2017+\dfrac{2017}{2}+\dfrac{2017}{2^2}+\dfrac{2017}{2^3}+...+\dfrac{2017}{2^{2017}}\\ =2017\cdot S\\ =2017\cdot\left(2-\dfrac{1}{2^{2017}}\right)\\ =4034-\dfrac{2017}{2^{2017}}\)