tìm n thỏa mãn 2+2^2+3.2^3 +4.2^4 +....+n.2^n =2^n+5
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\(Đặt\) \(A=2.2^2+3.2^3+4.2^4+...+n.2^n\)
\(2A=2.2^3+3.2^4+4.2^5+....+n.2^{n+1}\)
\(2A-A=2.2^3+3.2^4+4.2^5+....+n.2^{n+1}-\left(2.2^2+3.2^3+4.2^4+...+n.2^n\right)\)
\(=-2.2^2-2^3-2^4-...-2^n+n.2^{n+1}\)
\(=-2^2-\left(2^2+2^3+...+2^n\right)+n.2^{n+1}\)
\(=-2^2-\left(2^{n+1}-2^2\right)+n.2^{n+1}\)
\(=\left(n-1\right).2^{n+1}\)
=> \(\left(n-1\right).2^{n+1}=2^{n+16}=2^{n+1}.2^{15}\)
\(\Leftrightarrow n-1=2^{15}\)
\(\Leftrightarrow n=2^{15}+1\)
\(A=2.2^2+3.2^3+...+n.2^n\)
\(2A=2.2^3+3.2^4+4.2^5+...+n.2^{n+1}\)
\(2A-A=\left(2.2^3+3.2^4+...+n.2^{n+1}\right)-\left(2.2^2+3.2^3+...+n.2^n\right)\)
\(A=-2.2^2-2^3-2^4-...-2^n+n.2^{n+1}\)
\(A=-2^2-\left(2^2+2^3+2^4+...+2^n\right)+n.2^{n+1}\)
\(A=-2^2-\left(2^{n+1}-2^2\right)+n.2^{n+1}\)
\(A=\left(n-1\right)2^{n+1}=\left(2n-2\right).2^n\)
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 \(A=2.2^2+3.2^3+4.2^4+...+n.2^n\)
=>\(2.A=2.2^3+3.2^4+4.2^5+...+n.2^{n+1}\)
=>\(A-2A=2.2^2+3.2^3+4.2^4+...+n.2^n-2.2^3-3.2^4-4.2^5-...-n.2^{n+1}\)
=>\(-A=2.2^2+\left(3.2^3-2.2^3\right)+\left(4.2^4-3.2^4\right)+...+\left(n.2^n-\left(n-1\right).2^n\right)-n.2^{n+1}\)
=>\(-A=2^3+2^3+2^4+...+2^n-n.2^{n+1}\)
=>\(-A=2^3+\left(2^3+2^4+...+2^n\right)-n.2^{n+1}\)
=>\(A=n.2^{n+1}-2^3-\left(2^3+2^4+...+2^n\right)\)
Đặt \(B=2^3+2^4+...+2^n\)
=>\(2.B=2^4+2^5+...+2^{n+1}\)
=>\(2.B-B=2^4+2^5+...+2^{n+1}-2^3-2^4-...-2^n\)
=>\(B=2^{n+1}-2^3\)
Lại có:\(A=n.2^{n+1}-2^3-\left(2^3+2^4+...+2^n\right)\)
=>\(A=n.2^{n+1}-2^3-B\)
=>\(A=n.2^{n+1}-2^3-\left(2^{n+1}-2^3\right)\)
=>\(A=n.2^{n+1}-2^3-2^{n+1}+2^3\)
=>\(A=n.2^{n+1}-2^{n+1}\)
=>\(A=\left(n-1\right).2^{n+1}\)
Mà \(A=2.2^2+3.2^3+4.2^4+...+n.2^n=2^{n+10}\)
=>\(\left(n-1\right).2^{n+1}=2^{n+10}\)
=>\(n-1=2^{n+10}:2^{n+1}\)
=>\(n-1=2^{n+10-n-1}\)
=>\(n-1=2^9\)
=>\(n-1=512\)
=>\(n=513\)
Vậy n=513
dài thế hình như cô giáo lớp mình giải còn ngắn hơn thế này