Đặt: \(A=2\cdot2^2+3\cdot2^3+...+n\cdot2^n\)
\(2A=2\cdot\left(2\cdot2^2+3\cdot2^3+4\cdot2^4+...+n\cdot2^n\right)\)
\(2A=2\cdot2^3+3\cdot2^4+4\cdot2^5+...+n\cdot2^{n+1}\)
\(2A-A=\left(2\cdot2^3+3\cdot2^4+...+n\cdot2^{n+1}\right)-\left(2\cdot2^2+3\cdot2^3+...+n\cdot2^n\right)\)
\(A=-2\cdot2^2+\left(2\cdot2^3-3\cdot2^3\right)+\left(3\cdot2^4-4\cdot2^4\right)+...+\left[\left(n-1\right)\cdot2^n-n\cdot2^n\right]+n\cdot2^{n+1}\)
\(A=-2\cdot2^2-2^3-2^4-...-2^n+n\cdot2^{n+1}\)
\(A=-2\left(2^2+2^2+2^3+...+2^{n-1}\right)+n\cdot2^{n+1}\)
Mà: \(2^2+2^3+...+2^{n-1}=2^n-2^2\)
\(A=-2\cdot\left(2^2+2^n-2^2\right)+n\cdot2^{n+1}\)
\(A=-2\cdot2^n+n\cdot2^{n+1}\)
\(A=2^{n+1}\left(n-1\right)\)
Ta có: \(2\cdot2^2+3\cdot2^3+...+n\cdot2^n=2^{n+11}\)
\(\Rightarrow2^{n+1}\cdot\left(n-1\right)=2^{n+11}\)
\(\Rightarrow n-1=2^{n+11}:2^{n+1}\)
\(\Rightarrow n-1=2^{n+11-n-1}\)
\(\Rightarrow n-1=2^{10}\)
\(\Rightarrow n-1=1024\)
\(\Rightarrow n=1025\)
