A = 2.2² + 3.2³ + 4.2⁴ + ... + n.2ⁿ 

2.A = 2.2³ + 3.2⁴ + 4.2⁵ + ...+ n.2ⁿ⁺¹

=> A - 2.A = 2.2² + (3.2³ - 2.2³)  + (4.2⁴ - 3.2⁴) + ...+ (n - n + 1).2ⁿ - n.2ⁿ⁺¹

=> A = 2.2² + 2³ + 2⁴ + ..+ 2ⁿ - n.2^{n+ 1}  = 2² + (2² + 2³ + ....+ 2^{n+ 1}) - (n+1).2ⁿ⁺¹

=> A =  - 2² -  (2² + 2³ + ....+ 2^{n+ 1}) + (n+1).2ⁿ⁺¹

Tính B = 2² + 2³ + ....+ 2^{n+ 1} => 2.B =  2³ + ....+ 2^{n+ 1} + 2ⁿ⁺² => 2B - B = 2ⁿ⁺² - 2² => B = 2ⁿ⁺² - 2²

Vậy A = 2² - 2ⁿ⁺² + 2² + (n+1).2ⁿ⁺¹ = (n+1).2ⁿ⁺¹ - 2^{n+ 2} = 2ⁿ⁺¹.(n + 1 - 2) = (n-1).2ⁿ⁺¹ = 2(n-1).2ⁿ

Theo bài cho  A = 2(n-1).2ⁿ = 2ⁿ⁺¹⁰ => 2(n - 1) = 2¹⁰ => n - 1 = 2⁹  = 512 => n = 513

Vậy.............

n= 513, tui chỉ biết đáp án nhưng không biết cách làm

đặt A=2+2^2+2^3+...+2^n

     2A=2^2+2^3+2^4+...+2^n+1 (1)

  2A-A=2\(^{n+1}\)-2

     A=2\(^{n+1}\)-2  (2)

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

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

                              ..............................

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

cộng vế với vế ta có 

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

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

     2+2\(^{n+10}\)=n.2\(^{n+1}\)-2\(^{n+1}\)+2

            2\(^{n+10}\)=2\(^{n+1}\).(n-1)

             2\(^{n+1}\). 2\(^9\)=2\(^{n+1}\).(n-1)

=>n-1=2\(^9\)

=>n=2^9+1=513

vậy n=513

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

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

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

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

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

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

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

Từ đây phương trình ban đầu tương đương với: 

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

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

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

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

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

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

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

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

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

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

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

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

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

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 \(n=2^{33}+1\)

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