已知数列an满足a1=1╱4an=an-1(-1)n╱an-1-2设bn=1╱an2,求数列bn的前n项和设cn=ansin(2n-1已知数列{an}满足a1=1/4,an=an-1/[(-1)∧n·an-1-2](n≥2,n∈N)（1）求数列{an}的通项公式（2）设bn=1/an²,求数列bn的前

Question

已知数列an满足a1=1╱4an=an-1(-1)n╱an-1-2设bn=1╱an2,求数列bn的前n项和设cn=ansin(2n-1
已知数列{an}满足a1=1/4 ,an=an-1/[(-1)∧n·an-1-2] (n≥2,n∈N) （1）求数列{an}的通项公式（2）设bn=1/an²,求数列bn的前n项和sn（3）设cn=ansin[（2n-1）π/2],数列cn的前n项和为Tn,求证：对于任意的n∈N,Tn＜4/7

Answer 1

a(n+1) = a(n)/[(-1)^(n+1)a(n)-2],
若a(n+1)=0,则a(n)=0,...,a(1)=0,与a(1)=1/4矛盾.因此,a(n)不为0.
1/a(n+1) = [(-1)^(n+1)a(n)-2]/a(n) = (-1)^(n+1) - 2/a(n) = -(-1)^n -2/a(n)
= (-2)*(-1)^n + (-1)^n - 2/a(n),
1/a(n+1) - (-1)^n = -2/a(n) -2*(-1)^n,
1/a(n+1) + (-1)^(n+1) = -2[1/a(n) + (-1)^n],
{1/a(n) + (-1)^n}是首项为1/a(1) + (-1) = 3,公比为-2的等比数列.
1/a(n) + (-1)^n = 3(-2)^(n-1),
1/a(n) = (-1)^(n-1) + 3(-2)^(n-1)
b(n) = 1/[a(n)]^2 = [(-1)^(n-1) + 3(-2)^(n-1)]^2 = 1 + 9*4^(n-1)+ 6*2^(n-1),
s(n) = b(1)+b(2)+...+b(n) = n + 9[4^n - 1]/(4-1) + 6[2^n - 1]/(2-1)
= n + 3[4^n - 1] +6[2^n - 1]
= n - 9 + 3*4^n + 6*2^n
1/a(n) = (-1)^(n-1) + 3(-2)^(n-1),
1/a(2n) = (-1)^(2n-1) + 3(-2)^(2n-1) = -3*2^(2n-1) - 1 = -6*4^(n-1) - 1,
a(2n) = -1/[6*4^(n-1) + 1].
1/a(2n-1) = (-1)^(2n-2) +3(-2)^(2n-2) = 1 + 3*4^(n-1),
a(2n-1) = 1/[1+3*4^(n-1)].
c(n) = a(n)sin[(2n-1)PI/2].
c(2n) = a(2n)sin[(4n-1)PI/2] = a(2n)sin[-PI/2] = -a(2n) = 1/[6*4^(n-1) + 1].
c(2n-1) = a(2n-1)sin[(4n-3)PI/2] = a(2n-1)sin[-3PI/2] = a(2n-1) = 1/[1+3*4^(n-1)].
c(2n-1)+c(2n) = 1/[1+3*4^(n-1)] + 1/[6*4^(n-1)+1] < 1/[3*4^(n-1)] + 1/[6*4^(n-1)]
= (1/2)(1/4)^(n-1),
t(n) = c(1)+c(2)+...+c(n).
t(2n) = c(1)+c(2) + c(3)+c(4) + ...+ c(2n-1)+c(2n)
< (1/2)[1 + 1/4 + ...+ (1/4)^(n-1)]
= (1/2)[1 - (1/4)^n]/(1-1/4)
= (2/3)[1- (1/4)^n]
< 2/3
t(2n-1) = t(2n) - c(2n) = t(2n) - 1/[6*4^(n-1)+1] < t(2n) < 2/3.
因此,总有,
t(n) < 2/3