Question

数列An和Bn中，An前n项和为Sn，点（n,Sn）在y=-x2+4x上，(n,Bn)在y=2的x次方上，求An通项公是。求（AnBn）

Answer 1

由题得：Sn=-n^2+4n, Bn=2^n
a1=S1=-1+4=3
n>=2:
an=Sn-S(n-1)=-n^2+4n-[-(n-1)^2+4(n-1)]=-2n+1+4=-2n+5
a1=-2*1+5=3,符合
故an=-2n+5
AnBn=(-2n+5)*2^n
和Tn=3*2+1*2^2+(-1)*2^3+.....(-2n+5)*2^n
2Tn=3*2^2+1*2^3+(-1)*2^4+....+(-2n+5)*2^(n+1)
相减得：Tn=3*2+2(2^2+2^3+...+2^n)+(-2n+5)*2^(n+1)
=6+2*4*(2^(n-1)-1)/(2-1)+2(-2n+5)*2^n
=6+4*2^n-8+(-4n+10)*2^n
=(-4n+14)*2^n-2