Question

设数列{an}是等差数列,且a1=1,a2-a1,a3-a1,a4+a1成等比数列

设数列{an}是等差数列,且a1=1,a2-a1,a3-a1,a4+a1成等比数列

(1)求数列an的前n项和

(2)设bn=a2n-1，求数列{bn}的前n项和

Answer 1

a(n) = 2q^(n-1),q>0.
12 = a(3) - a(2) = 2q^2 - 2q,
0 = q^2 - q - 6 = (q-3)(q+2),
q = 3.
a(n) = 2*3^(n-1).
b(n) = 1 + 2(n-1) = 2n - 1.
s(n) = a(1)+a(2)+...+a(n) + b(1)+b(2)+...+b(n)
= 2[3^n - 1]/(3-1) + n(n+1) - n
= 3^n - 1 + n^2