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

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 $n=1025$