Question

[Nguyen Ngoc Hoa]

Tính S = 1 + 2 + \(2^2\) +\(2^3\) + \(2^4\) + ...+ \(2^{98}\) 

và chứng minh S chia hết cho 7

Answer 1

[o0o I am a studious person o0o]

\(S=1+2+2^2+2^3+2^4+.....+9^{98}\)

\(\Rightarrow2S=2+2^2+2^4+...+2^{99}\)

\(\Rightarrow S=2S-S=1-2^{99}\)

Answer 2

[Le Thi Khanh Huyen]

\(S=1+2+2^2+2^3+...+2^{98}\)

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

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

\(=\left(1+2^3+2^6+...+2^{93}+2^{96}\right)\left(1+2+2^2\right)\)

\(=\left(1+2^3+2^6+...+2^{93}+2^{96}\right).7\)chia hết cho 7.

Vậy S chia hết cho 7.