已知数列{an}满足an=2a(n-1)+2^n-1(n∈N+,且n>=2),a4=81
1个回答
an=2a(n-1)+2^n-1
an/2^n=a(n-1)/2^(n-1) +1-1/2^n
an/2^n-a(n-1)/2^(n-1)=1-1/2^n
a2/2^2-a1/2=1-1/2^2
a3/2^3-a2/2^2=1-1/2^3
...
an/2^n-a(n-1)/2^(n-1)=1-1/2^n
左右两边分别相加：
an/2^n-a1/2=(n-1)-[1/2^2+..+1/2^n]
an/2^n=a1/2+(n-1)-1/2^2*[1-1/2^(n-1)]/(1-1/2)
=a1/2+(n-1)-(1/2-1/2^n)
=a1/2+n-3/2+1/2^n
a1=5 (a4求a3-a2-a1,
an=(n+1)*2^n+1
Sn=a1+a2+..+an=n+2^1+2^2+..+2^n+1*2^1+2*2^2+..+n*2^n
=n+2*(1-2^n)/(1-2)+1*2^1+2*2^2+..+n*2^n
bn=1*2^1+2*2^2+..+n*2^n
2bn=1*2^2+.+(n-1)*2^n+n*2^(n+1)
相减：
-bn=2+2^2+.+2^n-n*2^(n+1)
=2*(1-2^n)/(1-2)-n*2^(n+1)
bn=-2*(1-2^n)/(1-2)+n*2^(n+1)
Sn=n+n*2^(n+1)
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