Question

在数列{An}中,A1=1,An+1=2An+2^n.(n+1为角标.)求数列{An}的前n项和Sn.

Answer 1

a(n+1)=2an+2^n

等式两边同除以2^n

a(n+1)/2^n=an/2^(n-1) +1

a(n+1)/2^n -an/2^(n-1)=1,为定值.

a1/2^(1-1)=1/1=1

数列{an/2^(n-1)}是以1为首项,1为公差的等差数列.

an/2^(n-1)=1+n-1=n

an=n×2^(n-1)

Sn=a1+a2+...+an=1×2^0+2×2^1+3×2^2+...+n×2^(n-1)

2Sn=1×2^1+2×2^2+...+(n-1)×2^(n-1)+n×2^n

Sn-2Sn=-Sn=2^0+2^1+2^2+...+2^(n-1)-n×2^n=(2^n -1)/(2-1) -n×2^n=(1-n)×2^n -1

Sn=(n-1)×2^n +1

^表示指数,用错位相减法.