Question

数列an满足a1=2,an+1=2an+2^n,求an和sn？

Answer 1

a(n+1)=2an+2ⁿ
等式两边同除以2ⁿ
a(n+1)/2ⁿ=an/2^(n-1) +1
a(n+1)/2ⁿ -an/2^(n-1)=1,为定值
a1/2^0=2/1=2,数列{an/2^(n-1)}是以2为首项,1为公差的等差数列
an/2^(n-1)=2+1×(n-1)=n+1
an=(n+1)×2^(n-1)
n=1时,a1=(1+1)×1=2,同样满足通项公式
数列{an}的通项公式为an=(n+1)×2^(n-1)
Sn=a1+a2+...+an=2×1+3×2+4×2²+...+(n+1)×2^(n-1)
2Sn=2×2+3×2²+...+n×2^(n-1)+(n+1)×2ⁿ
Sn-2Sn=-Sn=2+2+2²+...+2^(n-1) -(n+1)×2ⁿ
=1+2+2²+...+2^(n-1) -(n+1)×2ⁿ +1
=1×(2ⁿ-1)/(2-1)-(n+1)×2ⁿ +1
=-n×2ⁿ
Sn=n×2ⁿ,6,