Question

[Mai Hiệp Đức]

Tìm số tự nhiên n thỏa mãn \(2\cdot2^2+3\cdot2^3+...n\cdot2^n=2^{n+5}\)

Answer 1

[Inequalities]

Đặt \(A=2.2^2+3.2^3+...+n.2^n\)

\(\Rightarrow2A=2.2^3+3.2^4+...+n.2^{n+1}\)

\(\Rightarrow A-2A=\)\(2.2^2+3.2^3+...+n.2^n\)\(-2.2^3-3.2^4-...-n.2^{n+1}\)

\(\Rightarrow-A=2.2^2+2^3+2^4+...+2^n-n.2^{n+1}\)

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

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

Đặt \(K=\left(2^2+2^3+2^4+...+2^{n+1}\right)\)

\(2K=\left(2^3+2^4+2^5+...+2^{n+2}\right)\)

\(2K-K=\left(2^3+2^4+2^5+...+2^{n+2}\right)\)\(-\left(2^2+2^3+2^4+...+2^{n+1}\right)\)

\(K=2^{n+2}-2^2\)

\(\Rightarrow A=-2^2-2^{n+2}+2^2+\left(n+1\right).2^{n+1}\)

\(\Rightarrow A=\left(n+1\right).2^{n+1}-2^{n+2}\)

\(\Rightarrow A=2^{n+1}\left(n+1-2\right)\)

\(\Rightarrow A=2^{n+1}\left(n-1\right)=2^{n+5}\Rightarrow2^4=n-1\Rightarrow n=17\)