\(2A=2.2^3+3.2^4+4.2^5+...+100.2^{101}\)

=> \(2A-A=100.2^{101}-\left(2^{100}+2^{99}+...+2^4+2^3\right)-2.2^2\)

Đặt \(B=2^3+2^4+...+2^{100}\Rightarrow2B=2^4+2^5+...+2^{101}\)

=> \(2B-B=2^{101}-2^3\Rightarrow B=2^{101}-2^3\)

=> \(2A-A=100.2^{101}-\left(2^{101}-2^3\right)-2.2^2\)

=> \(A=\left(100.2^{101}-2^{101}\right)+2^3-2^3\)=\(99.2^{101}\)

helllo

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\(a,A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-..-\frac{1}{3.2}-\frac{1}{2.1}\)

\(A=\frac{1}{100}-\left(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\right)\)

\(A=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(A=\frac{1}{100}-1+\frac{1}{100}\)

\(A=\frac{2}{100}-1\)

\(A=\frac{1}{50}-1\)

\(A=\frac{-49}{50}\)

b,\(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n=2^{n+34}\)        (1)

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

\(\Rightarrow2B=2.\left(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n\right)\)

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

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

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

             \(B=-2^3-2^4-2^5-...-2^{n+1}-2.2^2\)

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

Đặt \(C=2^3+2^4+2^5+2^n\)

\(\Rightarrow2C=2.(2^3+2^4+2^5+...+2^n)\)

         \(C=2^4+2^5+2^6+...+2^{n+1}\)

\(2C-C=\left(2^4+2^5+2^6+...+2^{n+1}\right)-\left(2^3+2^4+2^5+...+2^n\right)\)

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

Khi đó :  \(B=-(2^{n+1}-2^3)+n.2^{n+1}-2^3\)

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

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

Vậy từ (1) ta có:\(\left(n-1\right),2^{n+1}=2^{n+34}\)

                           \(2^{n+34}-\left(n-1\right).2^{n+1}=0\)

                          \(2^{n+1}.[2^{33}-\left(n-1\right)]=0\)

Do đó \(2^{33}-n+1=0\)( Vì \(2^{n+1}\ne0\)với mọi \(n\))

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

Vậy \(n=2^{33}+1\)

Đặt \(A=2.2^2+3.2^3+...+n.2^n\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Lời giải:

$2^n+34=2.2^2+3.2^3+....+n.2^n$

$2^{n+1}+68=2.2^3+3.2^4+....+n.2^{n+1}$

Trừ theo vế:

$2^n+34=n.2^{n+1}-(8+2^3+2^4+...+2^n)$

$n.2^{n+1}-2^n-42=2^3+2^4+...+2^n$

$n.2^{n+2}-2^{n+1}-84=2^4+....+2^{n+1}$

Trừ theo vế:

$n.2^{n+1}-2^n-42=2^{n+1}-8$

$2^n(2n-3)=34=17.2$

$\Rightarrow 2^n=2$ và $2n-3=17$ (vô lý)

Vậy không tìm được $n$.