等差数列{an}{bn}前n项和Sn Tn,已知Sn/Tn=2n/（3n+1）,求a100/b100
来源：学生作业帮助网 编辑：六六作业网 时间：2024/11/22 05:59:30
等差数列{an}{bn}前n项和Sn Tn,已知Sn/Tn=2n/（3n+1）,求a100/b100
等差数列{an}{bn}前n项和Sn Tn,已知Sn/Tn=2n/（3n+1）,求a100/b100
等差数列{an}{bn}前n项和Sn Tn,已知Sn/Tn=2n/（3n+1）,求a100/b100
有等差数列的性质：a199+a1=2(a100),同理b199+b1=2(b100),
{an}的前199项和为S199=（a1+a199）*199/2,
{bn}的前199项和为T199=（b1+b199）*199/2,
S199/T199=a100/b100=(2*199)/(3*199+1)=199/299
1用特殊值法当n＝1时知a1＝ b1=
a(n) = a + (n-1)c. S(n) = na + n(n-1)c/2.
b(n) = b + (n-1)d. T(n) = nb + n(n-1)d/2.
2n/(3n+1) = S(n)/T(n) = [na + n(n-1)c/2]/[nb + n(n-1)d/2]
= [a + (n-1)c/2]/[b + (n-1)d/2],
= [2a + (...
全部展开
a(n) = a + (n-1)c. S(n) = na + n(n-1)c/2.
b(n) = b + (n-1)d. T(n) = nb + n(n-1)d/2.
2n/(3n+1) = S(n)/T(n) = [na + n(n-1)c/2]/[nb + n(n-1)d/2]
= [a + (n-1)c/2]/[b + (n-1)d/2],
= [2a + (n-1)c]/[2b + (n-1)d]
2/4 = S(1)/T(1) = a/b.
b = 2a.
4/7 = S(2)/T(2) = [2a + c]/[2b + d] = [2a + c]/[4a + d],
4[4a+d] = 7[2a+c],
16a+4d = 14a + 7c,
d = (7c - 2a)/4
6/10 = S(3)/T(3) = [2a + 2c]/[2b + 2d] = (a+c)/(b+d)
5(a+c) = 3[b + d] = 3[2a + (7c-2a)/4] = 6a + 21c/4 - 3a/2,
a/2 = c/4,
c = 2a.
d = (7c - 2a)/4 = (14a - 2a)/4 = 12a/4 = 3a.
a(100) = a + 99c = a + 2*99a,
b(100) = b + 99d = 2a + 3*99a
a(100)/b(100) = [a + 2*99a]/[2a + 3*99a] = [1 + 2*99]/[2 + 3*99]
= 199/299
收起
求S199/T199 即可