- 已知数列{an}相邻的两项a[2k
- 已知数列{an}相邻的两项a[2k-1],a[2k]是关于x的方程x^2-(3k+2^k)*x+3k*2^k=0的两个根,且a[2k-1]<=a[2k],(k=1,2,3...)
(1)求a1,3,a5,a7及a[2n] ,(n>=4)(不必证明)
(2)求数列{an}的前2n项和s[2n]
- (1)
(x-2^k)(x-3k)=0
因为
3k>2^k (k<=3)
3k<2^k (k>3)
a1=2,a2=3
a3=4,a4=6
a5=8,a6=9
a7=12,a8=16
a(2k)=2^k
(2)
S(2n)
=(3*1+3*2+3*3+.....3*n)+(2^1+2^2+......2^n)
=3/2*n*(n-1)+2*(2^n-1)