Question

已知数列{An}为公差不为零的等差数列,a1=2,且a2,a3,a4+1成等比数列,。

已知数列{An}为公差不为零的等差数列,a1=2,且a2,a3,a4+1成等比数列,设bn=2/n(an+2)，求数列{bn}的前n项和Sn。

Answer 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
bn=an+2^an
= 2n+ 2^(2n)
Sn =b1+b2+...+bn
= n(n+1) + (4/3)(2^(2n)-1 )
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Answer 2

a4^2=a2a8
(a1+3d)^2=(a1+d)(a1+7d)
2a1d-2d^2=0
a1=2代入，整理，得
d(d-2)=0
d=0(已知公差不为0，舍去)或d=2
an=a1+(n-1)d=2+2(n-1)=2n
数列{an}的通项公式为an=2n
3^(an)=3^(2n)=9^n
sn=a1+a2+...+an=9^1+9^2+...+9^n=9(9^n-1)/(9-1)=9^(n+1)
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9/8