Question

已知数列{an}中,a1=5且an=2an-1+2n-1（n≥2且n∈N+）

Answer 1

n≥2时,

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

等式两边同除以2^n

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

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

(an -1)/2^n -[a(n-1) -1]/2^(n-1)=1,为定值.

(a1-1)/2=(5-1)/2=2,数列{(an -1)/2}是以2为首项,1为公差的等差数列.

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

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

n=1时,a1=4+1=5,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

=2×2+3×2^2+...+(n+1)×2^n +n

令Cn=2×2+3×2^2+...+(n+1)×2^n,

则2Cn=2×2^2+3×2^3+...+n×2^n+(n+1)×2^(n+1)

Cn-2Cn=-Cn=4+2^2+2^3+...+2^n -(n+1)×2^(n+1)

=1+2+2^2+...+2^n -(n+1)×2^(n+1) +1

=1×[2^(n+1) -1]/(2-1) -(n+1)×2^(n+1) +1

=-n×2^(n+1)

Cn=n×2^(n+1)

Sn=Cn +n=n×2^(n+1) +n