已知等差数列an的公差d大于0,且a3,a5是方程x^2-14x+45=0的两根求数列an的通项公式记bn=2的an次方+你,求数列bn的前n项和Sn

d>0，a(n+1)-an=d>0，数列单调递增
x²-14x+45=0
(x-5)(x-9)=0
x=5或x=9
数列单调递增，a3a3=5 a5=9
a5-a3=2d=9-5=4
d=2
a1=a3-2d=5-2×2=1
an=a1+(n-1)d=1+2(n-1)=2n-1
数列{an}的通项公式为an=2n-1
bn=2^an +n=2^(2n-1)+n=4ⁿ/2 +n
Sn=b1+b2+...+bn
=(1/2)(4+4²+...+4ⁿ) +(1+2+...+n)
=(1/2)×4×(4ⁿ-1)/(4-1) +n(n+1)/2
=(2/3)×(4ⁿ-1) +n(n+1)/2
=2^(2n+1) /3 +n(n+1)/2 -2/3

an =a1+(n-1)da3+a5= 14 2a1+6d=14a1+3d=7 (1)a3.a5=45(a1+2d)(a1+4d)=45(7-d)(7+d)=45d^2=4d=2from (1),a1=1an = 1+2(n-1) = 2n-1bn=2^(an) +n= 2^(2n-1) +nSn = b1+b2+...+bn= 2( 2^(2n) -1) /(4-1) + n(n+1)/2=(2...