Question

Tìm số TN n thỏa mãn điều kiện: \(2.2^2+3.2^3+4.2^4+....+n.2^n=2^{n+11}\)

Answer 1

Đặt A=\(2.2^2+3.2^2+...+n.2^n\)

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

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

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

Đặt \(B=2^3+2^4+...+2^n\Rightarrow2B=2^4+2^5+...+2^{n+1}\)

\(\Rightarrow2B-B=\left(2^4+2^5+...+2^{n+1}\right)-\left(2^3+2^4+...+2^n\right)\)

\(\Rightarrow B=2^{n+1}-2^3\)

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

=\(n.2^{n+1}-8-2^{n+1}+8=\left(n-1\right).2^{n+1}\)

Mà A=\(2^{n+11}\Rightarrow\left(n-1\right).2^{n+1}=2^{n+11}\)

\(\Rightarrow2^{n+1}.\left(n-1\right)=2^{n+1}.2^{10}\Rightarrow n-1=2^{10}=1024\Rightarrow n=2015\)

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