在数列｛an｝中,若a1=1,a（n+1）=2an+3（n≥1）,则该数列的通项an等于?

Question

在数列｛an｝中,若a1=1,a（n+1）=2an+3（n≥1）,则该数列的通项an等于?

Answer 1

由于:
a(n+1)=2an+3
则有:
[a(n+1)+3]=2(an+3)
[a(n+1)+3]/(an+3)=2
则:{an+3}为公比为2的等比数列
则:
an+3
=(a1+3)*2^(n-1)
=4*2^(n-1)
=2^2*2^(n-1)
=2^(n+1)
则:
an=2^(n+1)-3
则:
nan=n[2^(n+1)-3]
则:
Sn=1*a1+2*a2+...+n*an
=1*[2^2-3]+2*[2^3-3]+...+n*[2^(n+1)-3]
=[1*2^2+2*2^3+...+n*2^(n+1)]-3(1+2+...+n)
=[1*2^2+2*2^3+...+n*2^(n+1)]-3n(n+1)/2
设Tn=1*2^2+2*2^3+...+n*2^(n+1)
则:
2Tn=1*2^3+2*2^4+...+(n-1)*2^(n+1)+n*2^(n+2)
两式相减,得:
-Tn=1*2^2+1*2^3+...1*2^(n+1)-n*2^(n+2)
则:
Tn
=-[2^2+2^3+...+2^(n+1)]+n*2^(n+2)
=-[4*(1-2^n)/(1-2)]+n*2^(n+2)
=(n-1)*2^(n+2)+4
则:
Sn=(n-1)*2^(n+2)-3n(n+1)/2+4