Question

[Mai Trọng Gia Long]

Cho S = \(1\) + \(2^1\) + \(2^2\) + \(2^3\) + \(2^4\) + . . . + \(2^{119}\)

a ) CMR : S chia hết cho 5 ;                                   b ) Tìm n thuộc N để S + 1 = \(2^n\)

Answer 1

[Nguyễn Minh Đăng]

a) Ta có: 

\(S=1+2+2^2+...+2^{119}\)

\(S=\left(1+2+2^2+2^3\right)+\left(2^3+2^4+2^5+2^6\right)+...+\left(2^{116}+2^{117}+2^{118}+2^{119}\right)\)

\(S=\left(1+2+2^2+2^3\right)+2^3\cdot\left(1+2+2^2+2^3\right)+...+2^{116}\cdot\left(1+2+2^2+2^3\right)\)

\(S=15+15\cdot2^3+...+15\cdot2^{116}\)

\(S=15\cdot\left(1+2^3+...+2^{116}\right)\) chia hết cho 5

b) \(S=1+2+2^2+...+2^{119}\)

\(\Rightarrow2S=2+2^2+2^3+...+2^{120}\)

\(\Rightarrow2S-S=\left(2+2^2+...+2^{120}\right)-\left(1+2+...+2^{119}\right)\)

\(\Leftrightarrow S=2^{120}-1\)

\(\Leftrightarrow2^n=S+1=2^{120}\)

\(\Rightarrow n=120\)