作业帮高一数列题,求第二问过程或方法
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高一数列题,求第二问过程或方法
高一数列题,求第二问过程或方法
高一数列题,求第二问过程或方法
2s(n) + a(n) = 1, 1 = a(1) + 2s(1) = 3a(1), a(1) = 1/3.
2s(n+1) + a(n+1) = 1,
0 = 1-1 = 2[s(n+1)-s(n)] + a(n+1) - a(n) = 2a(n+1) + a(n+1) - a(n),
a(n+1) = a(n)/3,
{a(n)}是首项为a(1)=1/3,公比为1/3的等比数列.
a(n) = (1/3)(1/3)^(n-1) = 1/3^n,
2/b(n+1) = 1/b(n) + 1/b(n+2),
1/b(n+2) - 1/b(n+1) = 1/b(n+1) - 1/b(n),
{1/b(n+1) - 1/b(n)} 是首项为1/b(2) - 1/b(1) = 2-1 = 1,的常数数列.
1/b(n+1) - 1/b(n) = 1,
1/b(n+1) = 1/b(n) + 1,
{1/b(n)}是首项为1/b(1)=1,公差为1的等差数列.
1/b(n) = 1 + (n-1) = n,
b(n) = 1/n.
c(n) = a(n)/b(n) = n/3^n,
t(n) = c(1)+c(2)+c(3)+...+c(n-1)+c(n)
= 1/3 + 2/3^2 + 3/3^3 + ... + (n-1)/3^(n-1) + n/3^n,
3t(n) = 1 + 2/3 + 3/3^2 + ... + (n-1)/3^(n-2) + n/3^(n-1),
2t(n) = 3t(n) - t(n) = 1 + 1/3 + 1/3^2 + ... + 1/3^(n-1) - n/3^n
= [1 - 1/3^n]/(1-1/3) - n/3^n
= (3/2)[1 - 1/3^n] - n/3^n
= 3/2 - (3/2 + n)/3^n.
t(n) = c(1)+c(2)+...+c(n) = 3/4 - (3/4 + n/2)/3^n,
2s(n) + a(n) = 1,
s(n) = [1 - a(n)]/2 = [1 - 1/3^n]/2,
t(n) - s(n) = 3/4 - (3/4 + n/2)/3^n - [1-1/3^n]/2
= 1/4 - (3/4 + n/2 - 1/2)/3^n
= 1/4 - (1/4 + n/2)/3^n,
0 < t(n) - s(n) = 1/4 - (1/4 + n/2)/3^n,
0 < 1 - (1+2n)/3^n,
0 < 3^n - (1+2n),
3^n - 1 = (3-1)[3^(n-1) + 3^(n-2) + ... + 3 + 1] = 2[3^(n-1) + 3^(n-2) + ... + 3 + 1]
> 2[1 + 1 + ... + 1 + 1]
= 2n,
因此,总有,3^n - 1 > 2n,
3^n > 1 + 2n,
t(n) - s(n) = 1/4 - (1/4 + n/2)/3^n >0.
所以,正整数k的最小值为1.