关于递推关系3 5 6 7

关于递推关系3 5 6 7
关于递推关系
3 5 6 7

关于递推关系3 5 6 7
(2)
a[2] / a[1] = 1 / 3
a[3] / a[2] = 2 / 4
a[4] / a[3] = 3 / 5
...
a[n - 1] / a[n - 2] = (n - 2) / n
a[n] / a[n - 1] = (n - 1) / (n + 1)
将左边全部乘起来再乘上a[1]得
a[1] * (a[2] / a[1]) * ...* (a[n] / a[n - 1])
=a[1] / a[1] * a[2] / a[2] * ...* a[n - 1] / a[n - 1] * a[n]
=a[n] = 2 * (1 / 3) * (2 / 4) * (3 / 5) ...* (n - 2) / n * (n - 1) / (n + 1)
= 1 * 2 * 2 * 3 / 3 * 4 / 4 * 5 / 5 ...* (n - 1) / (n - 1) / n / (n + 1)
= 2 * 2 / n ／ (n + 1)
= 4 / [(n + 1) * n]
(3) //sqrt 表示根号下
1 / [sqrt(n + 1) + sqrt(n)]
={1 * [sqrt(n + 1) - sqrt(n)]} / {[sqrt(n + 1) + sqrt(n)] * [sqrt(n + 1) - sqrt(n)]}
=[sqrt(n + 1) - sqrt(n)] / [(n + 1) - n]
=[sqrt(n + 1) - sqrt(n)] / 1
=sqrt(n + 1) - sqrt(n)
所以 a[n] - a[n - 1] = sqrt(n + 1) - sqrt(n)
a[2] - a[1] = sqrt3 - sqrt2
a[3] - a[2] = sqrt4 - sqrt3
...
a[n] - a[n - 1] = sqrt(n + 1) - sqrt(n)
左边全部加起来
a[n] - a[n - 1] + a[n - 1] + a[n - 2] + ...+a[3] - a[2] + a[2] - a[1]
=a[n] - a[1] = sqrt(n + 1) - sqrt(2)
a[n] = sqrt(n + 1) - sqrt(2) + a[1] = sqrt(n + 1) - sqrt(2) + 1
(5)
因为 b[n + m] = b[n] * b[m]
b[2] = b[1] * b[1] = 2^2
b[3] = b[1] * b[2] = 2^3
所以 b[n] = b[n - 1] * b[1] = b[n - 2] * b[1]^2 = ...= b[1] * b[1]^(n - 1)
= b[1]^n = 2^n
(6)
a[n] - 2a[n - 1] = 1
2(a[n - 1] - 2a[n - 2]) = 2
2^2 * (a[n - 2] - 2a[n - 3]) = 2^2
2^3 * (a[n - 3] - 2a[n - 4]) = 2^3
.
2^(n - 2) * (a[2] - 2a[1]) = 2^(n - 2)
左边加起来得
a[n] - 2^(n - 1) * a[1] = 2^0 + 2^1 + 2^2 ...+ 2^(n - 2) = 2^(n - 1) - 1
a[n] = 2^(n - 1) + 2^(n - 1) - 1 = 2^n - 1
(7)
由 a[n + 1] - a[n] = 2^n
所以
a[n] - a[n - 1] = 2^(n - 1)
a[n - 1] - a[n - 2] = 2^(n - 2)
...
a[2] - a[1] = 2^1
左边加起来
a[n] - a[1] = 2^1 + 2^2 + 2^3 + ...+ 2^(n - 1) = 2^n - 1 - 1
a[n] = 2^n - 1 - 1 + a[1] = 2^n
S[n] = 2^1 + 2^2 + 2^3 + ...+ 2^n
= 1 - 1 + 2^1 + 2^2 + 2^3 ...+ 2^n
= 2^(n + 1) - 1 - 1
= 2^(n + 1) - 2