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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 \(K=\left(2^2+2^3+2^4+...+2^{n+1}\right)\)
\(2K=\left(2^3+2^4+2^5+...+2^{n+2}\right)\)
\(2K-K=\left(2^3+2^4+2^5+...+2^{n+2}\right)\)\(-\left(2^2+2^3+2^4+...+2^{n+1}\right)\)
\(K=2^{n+2}-2^2\)
\(\Rightarrow A=-2^2-2^{n+2}+2^2+\left(n+1\right).2^{n+1}\)
\(\Rightarrow A=\left(n+1\right).2^{n+1}-2^{n+2}\)
\(\Rightarrow A=2^{n+1}\left(n+1-2\right)\)
\(\Rightarrow A=2^{n+1}\left(n-1\right)=2^{n+5}\Rightarrow2^4=n-1\Rightarrow n=17\)
\(\Leftrightarrow2^n.\left(2^{-1}+4\right)=9.2^5\)
\(\Leftrightarrow2^n.4,5=4,5.2^6\)
\(\Rightarrow2^n=2^6\)
\(\Rightarrow n=6\)
\(\frac{2^n}{2}+2^2.2^n=9.2^5\)
=> \(2^{n-1}+2^{2+n}=2^5.9\)
=> \(2^{n-1}+2^{n-1+3}=2^5.9\)
=> \(2^{n-1}.\left(1+2^3\right)=2^5.9\)
=> \(2^{n-1}.9=2^5.9\)
=> \(2^{n-1}=2^5\)
=> \(n-1=5\)
Vậy n = 6.
Ta có:
\(\frac{2^n}{2}+2^2.2^n=2^{n-1}+2^3.2^{n-1}=2^{n-1}.\left(1+2^3\right)\)
\(=2^{n-1}.9=9.2^5\)
Chia cả 2 vế cho 9 thì có:
\(2^{n-1}=2^5\Rightarrow2^n=2^6\Rightarrow n=6\)
