Question

[ngothithoi]

cho a =1+4+4²+.......+4⁹⁸

a rút gọn a 

b chứng minh rằng achia hết cho 7 

Answer 1

[Phùng Minh Quân]

\(a)\) \(A=1+4+4^2+...+4^{98}\)

\(4A=4+4^2+4^3+...+4^{99}\)

\(4A-A=\left(4+4^2+4^3+...+4^{99}\right)-\left(1+4+4^2+...+4^{98}\right)\)

\(3A=4^{98}-1\)

\(A=\frac{4^{98}-1}{3}\)

Vậy \(A=\frac{4^{98}-1}{3}\)

\(b)\) \(A=1+4+4^2+...+4^{98}\)

\(A=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...+\left(4^{96}+4^{97}+4^{98}\right)\)

\(A=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...+4^{96}\left(1+4+4^2\right)\)

\(A=\left(1+4+16\right)+4^3\left(1+4+16\right)+...+4^{98}\left(1+4+16\right)\)

\(A=21+4^3.21+...+4^{98}.21\)

\(A=21\left(1+4^3+...+4^{98}\right)\)

\(A=7.3\left(1+4^3+...+4^{98}\right)⋮7\)

Vậy \(A⋮7\)

Chúc bạn học tốt ~ 

Answer 2

[Hoàng Ninh]

\(A=1+4+4^2+........+4^{98}\)

\(\Rightarrow4A=4+4^2+4^3+..........+4^{99}\)

\(\Rightarrow4A-A=\left(4+4^2+4^3+...........+4^{99}\right)-\left(1+4+4^2+.......+4^{98}\right)\)

\(\Rightarrow3A=4^{99}-1\)

\(\Rightarrow A=\frac{4^{99}-1}{3}\)

b, 

\(A=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...........+\left(4^{96}+4^{97}+4^{98}\right)\)

\(A=\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+............+4^{96}.\left(1+4+4^2\right)\)

\(A=21+4^3.21+..........+4^{96}.21\)

\(A=21.\left(1+4^3+............+4^{96}\right)\)

Mà 21 \(⋮7\)\(\Rightarrow21.\left(1+4^3+..........+4^{96}\right)⋮7\)

\(\Rightarrow A⋮7\)

Vậy A chia hết cho 7 ( ĐPCM )

P/s: Điều phải chứng minh nha bạn

Chúc bạn học tốt!