Tìm số tự nhiên n để 2.2^2+3.2^3+4.2^4+…+ (n-1).2^n-1+n.2^n = 2^n+11

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1. Đáp án:

   Vậy $n=1025$

   Giải thích các bước giải:

   Đặt $A = 2.2^2+3.2^3+4.2^4+…+ (n-1).2^{n-1}+n.2^n$

   $=>2A= 2.(2.2^2+3.2^3+4.2^4+…+ (n-1).2^{n-1}+n.2^n )$

   $=>2A=2.2^3+3.2^4+4.2^5+…+ (n-1).2^{n}+n.2^{n+1}$

   $=>2A-A=(2.2^3+3.2^4+4.2^5+…+ (n-1).2^{n}+n.2^{n+1})-(2.2^2+3.2^3+4.2^4+…+ (n-1).2^{n-1}+n.2^n)$

   $=>2A-1.A=(2.2^3-3.2^3)+(3.2^4-4.2^4)+…+ [(n-1).2^{n}-n.2^n]+(n.2^{n+1}-2.2^2)$

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

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

   Đặt $B=-2^3+(-2^4)+…+ (-2^n)$

   $2B=-2^4+(-2^5)+…+ (-2^{n+1})$

   $2B-B=[-2^4+(-2^5)+…+ (-2^{n+1})]-[-2^3+(-2^4)+…+ (-2^n)]$

   $B= (-2^{n+1})-(-2^3)$

   $B= -2^{n+1}+2^3$

   $=>A=(-2^{n+1}+2^3)+(n.2^{n+1}-2.2^2)$

   $=>A=-2^{n+1}+n.2^{n+1}+(2^3-2.2^2)$

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

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

   $=>(-1+n).(2^{n+1})=2^{n+11}$

   $=>-1+n=2^{n+11}:(2^{n+1})$

   $=>-1+n=2^{10}$

   $=>n=2^{10}+1$

   $=>n=1025$

   Vậy $n=1025$

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