\(Đặt\) \(A=2.2^2+3.2^3+4.2^4+...+n.2^n\)

\(2A=2.2^3+3.2^4+4.2^5+....+n.2^{n+1}\)

\(2A-A=2.2^3+3.2^4+4.2^5+....+n.2^{n+1}-\left(2.2^2+3.2^3+4.2^4+...+n.2^n\right)\)

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

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

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

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

=> \(\left(n-1\right).2^{n+1}=2^{n+16}=2^{n+1}.2^{15}\)

\(\Leftrightarrow n-1=2^{15}\)

\(\Leftrightarrow n=2^{15}+1\)

\(A=2.2^2+3.2^3+...+n.2^n\)

\(2A=2.2^3+3.2^4+4.2^5+...+n.2^{n+1}\)

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

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

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

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

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

Từ đây phương trình ban đầu tương đương với: 

\(\left(2n-2\right).2^n=2^{n+34}\)

\(\Leftrightarrow\left(2n-2\right).2^n=2^n.2^{34}\)

\(\Leftrightarrow n-1=2^{33}\)

\(\Leftrightarrow n=2^{33}+1\)

Đặt A = 2.2 2 + 3.2 3 + 4.2 4 + . . . + m .2 m Ta có: A = 2.2 2 + 3.2 3 + 4.2 4 + . . . + m .2 m ⇒ 2 A = 2 ( 2.2 2 + 3.2 3 + 4.2 4 + . . . + n .2 n ) ⇒ 2 A = 2.2 3 + 3.2 4 + 4.2 5 + . . . + m .2 m + 1 ⇒ 2 A − A = 2.2 2 + ( 3.2 3 − 2.2 3 ) + . . . + ( m − m + 1 ) .2 m − m .2 m + 1 ⇒ A = 2.2 2 + 2 3 + 2 4 + . . . + 2 n − n .2 n + 1 ⇒ A = 2 2 + ( 2 2 + 2 3 + . . . + 2 m + 1 ) − ( m + 1 ) .2 m + 1 ⇒ A = − 2 2 − ( 2 2 + 2 3 + . . . + 2 m + 1 ) + ( m + 1 ) .2 m + 1 Đặt B = 2 2 + 2 3 + . . . + 2 m + 1 ⇒ 2 B = 2 3 + 2 4 + . . . + 2 m + 2 ⇒ 2 B − B = 2 m + 2 − 2 2 ⇒ B = 2 m + 2 − 2 2 ⇒ A = 2 2 − 2 m + 2 + 2 2 + ( m + 1 ) .2 m + 1 ⇒ A = ( m + 1 ) .2 m + 1 − 2 m + 2 ⇒ A = 2 m + 1 ( m + 1 − 2 ) ⇒ A = ( m − 1 ) .2 m + 1 = 2 ( m − 1 ) .2 n Mà A = 2 ( m − 1 ) .2 m = 2 m + 10 ⇒ 2 ( m + 1 ) = 2 10 ⇒ m − 1 = 2 9 ⇒ m − 1 = 512 ⇒ m = 513 Vậy m = 513

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

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

Vậy...

p hk tốt nha