Câu hỏi của Nguyễn Thị Thu Hà

Question

S = 2+2.22+3.23 +... +2016.22016
          1) Chứng tỏ S+2013 chia hết cho 22017+1
           2) Tìm số dư khi chia S cho 8

Nguyễn Tuấn Tú · Accepted Answer

$S=2+2.2^2+3.2^3+...+2016.2^{2016}$

$2S=2^2+2.2^3+3.2^4+...+2016.2^{2017}$

$2S-S=S=\text{​​}\text{​​}\text{​​}\text{​​}2^2+2.2^3+3.2^4+...+2016.2^{2017}-2-2.2^2-3.2^3-...-2016.2^{2016}$

$S=2\left(0-1\right)+2^2\left(1-2\right)+2^3\left(2-3\right)+...+2^{2016}\left(2015-2016\right)+2^{2017}.2016$

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

Đặt $A=2+2^2+2^3+...+2^{2016}$

$2A=2^2+2^3+2^4+...+2^{2017}$

$2A-A=A=2^2+2^3+2^4+...+2^{2017}-2-2^2-2^3-...-2^{2016}$

$A=2^{2017}-2$

Thay vào S ta được:
$S=-2^{2017}+2+2^{2017}.2016$

$S=2^{2017}.2015+2$

Ta có $S+2013=2^{2017}.2015+2+2013$

$S+2013=2^{2017}.2015+2015$

$S+2013=2015\left(2^{2017}+1\right)$

Suy ra $S+2013⋮2^{2017}+1$

Vậy $S+2013⋮2^{2017}+1$ (đpcm)Câu trả lời: $S=2+2.2^2+3.2^3+...+2016.2^{2016}$