Câu hỏi của Phạm Thị Như Anh

Question

a) Tính S = $\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}$

b) Cho A = $\dfrac{1}{2017}$+ $\dfrac{2}{2017^2}$ + $\dfrac{3}{2017^3}$...

Mysterious Person · Accepted Answer

a) $S=\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}$

$\Rightarrow2S=\dfrac{2\left(2+2^2+2^3+...+2^{2017}\right)}{1-2^{2017}}$

$2S=\dfrac{2^2+2^3+2^4+...+2^{2018}}{1-2^{2017}}$

$\Rightarrow2S-S=S=\dfrac{2^2+2^3+2^4+...+2^{2018}}{1-2^{2017}}-\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}$

$S=\dfrac{\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)}{1-2^{2017}}$

$S=\dfrac{2^{2018}-2}{1-2^{2017}}=\dfrac{-2\left(1-2^{2017}\right)}{1-2^{2017}}=-2$ vậy $S=-2$

Câu trả lời:

a) $S=\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}$

$\Rightarrow2S=\dfrac{2\left(2+2^2+2^3+...+2^{2017}\right)}{1-2^{2017}}$

$2S=\dfrac{2^2+2^3+2^4+...+2^{2018}}{1-2^{2017}}$

$\Rightarrow2S-S=S=\dfrac{2^2+2^3+2^4+...+2^{2018}}{1-2^{2017}}-\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}$

$S=\dfrac{\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)}{1-2^{2017}}$

$S=\dfrac{2^{2018}-2}{1-2^{2017}}=\dfrac{-2\left(1-2^{2017}\right)}{1-2^{2017}}=-2$ vậy $S=-2$