Question

已知数列{an}是公差不为0的等差数列，a1=2，且a2,a3,a4+1成等比数列 (1)求数列的通项公式 (2)设bn=an+...

已知数列{an}是公差不为0的等差数列，a1=2，且a2,a3,a4+1成等比数列 (1)求数列的通项公式 (2)设bn=an+2^an，求数列{bn}的前n项和Sn

Answer 1

设a(2)=2+d,a(3)=2+2d,a(4)=2+3d,
则a(3)^2=a(2)*(1+a(4)),
4+8d+4d^2=(2+d)(3+3d)=6+9d+3d^2,
d^2-d-2=0,(d-2)(d+1)=0,则d=2,d=-1
则a(n)=2+(n-1)*2=2n或a(n)=2-(n-1)=1-n
当a(n)=2n时，b(n)=a(n)+2^(a(n))=2n+2^(2n),
sn=2*(1+2+3+...+n)+4+4^2+4^3+...+4^n=2*n*(n+1)/2+4(4^n-1)/(4-1)=n(n+1)+4/3*(4^n-1)
当a(n)=1-n时，b(n)=1-n+2^(1-n),
sn=n-(1+2+3+...+n)+2^0+2^(-1)+2^(-2)+...+2^(1-n)
=-(1+2+3+...+(n-1))+1+2^(-1)+2^(-2)+...+2^(1-n)
=-(n-1)n/2+(1-2^(1-n))/(1-1/2)
=n(1-n)/2+2*(1-2^(1-n))

Answer 2

(1)
an=2+(n-1)d
a2,a3,a4+1成等比数列

a2(a4+1)=(a3)^2
(2+d)(3+3d)=(2+2d)^2
6+9d+3d^2 = 4d^2+8d+4
d^2-d-2=0
d=2 or -1(rejected)
an = 2+(n-1)2 = 2n
(2)
bn=an+2^an
= 2n+ 2^(2n)
Sn =b1+b2+...+bn
= n(n+1) + (4/3)(2^(2n)-1 )

Answer 3

1..a2,a3,a4+1 等比, (a3)^2=a2*(a4+1) ∴ (2+2d)^2=(2+d)*(2+3d+1) 解得公差d=2
∴an=a1+(n-1)d=2n
2..bn=2n+2^2n
Sn=((1+2n)n/2)+4（1-4^n)/(1-4)
=(n+2n^2)/2+4(4^n-1)/3