Câu hỏi của trịnh bảo trung

Question

chứng minh A = 2 + 2² + 2³ + … + 2

Nguyễn Lê Phước Thịnh · Accepted Answer

Ta có: $A=2+2^2+2^3+\cdots+2^{120}$

$=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\cdots+\left(2^{118}+2^{119}+2^{120}\right)$

$=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+\cdots+2^{118}\left(1+2+2^2\right)$

$=7\left(2+2^4+\cdots+2^{118}\right)$ ⋮7

Ta có: $A=2+2^2+2^3+\cdots+2^{120}$

$=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+\cdots+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)$

$=2\left(1+2+\cdots+2^4\right)+2^6\left(1+2+\cdots+2^4\right)+\cdots+2^{116}\left(1+2+\cdots+2^4\right)$

$=31\left(2+2^6+\ldots+2^{116}\right)$ ⋮31

Ta có: $A=2+2^2+2^3+\cdots+2^{120}$

$=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^{119}+2^{120}\right)$

$=2\left(1+2\right)+2^3\left(1+2\right)+\cdots+2^{119}\left(1+2\right)=3\left(2+2^3+\cdots+2^{119}\right)$ ⋮3

Ta có: A⋮3

A⋮7

mà ƯCLN(3;7)=1

nên A⋮3*7

=>A⋮21

Câu trả lời:

Ta có: $A=2+2^2+2^3+\cdots+2^{120}$

$=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\cdots+\left(2^{118}+2^{119}+2^{120}\right)$

$=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+\cdots+2^{118}\left(1+2+2^2\right)$

$=7\left(2+2^4+\cdots+2^{118}\right)$ ⋮7

Ta có: $A=2+2^2+2^3+\cdots+2^{120}$

$=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+\cdots+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)$

$=2\left(1+2+\cdots+2^4\right)+2^6\left(1+2+\cdots+2^4\right)+\cdots+2^{116}\left(1+2+\cdots+2^4\right)$

$=31\left(2+2^6+\ldots+2^{116}\right)$ ⋮31

Ta có: $A=2+2^2+2^3+\cdots+2^{120}$

$=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^{119}+2^{120}\right)$

$=2\left(1+2\right)+2^3\left(1+2\right)+\cdots+2^{119}\left(1+2\right)=3\left(2+2^3+\cdots+2^{119}\right)$ ⋮3

Ta có: A⋮3

A⋮7

mà ƯCLN(3;7)=1

nên A⋮3*7

=>A⋮21