Python等比数列

等比数列

• 公比用q表示，$q\ne 0$
  $n\in N^{\ast }$
  $a_1\ne 0$，等比数列中每一项都不等于0
• 通项公式：$a_n=a_1\cdot q^{n-1}$
• 等比数列求和：
  $S_n=n\cdot a_1(q=1)$
  $S_n=\frac{a_1(1-q^n)}{1-q}(q\ne 1)$

def conditions(a1, q, n):
    assert a1 != 0
    assert q != 0
    assert isinstance(n, int)
    assert n > 0


def a_n(a1, q, n):
    conditions(a1, q, n)
    return a1 * q ** (n - 1)


def s_n(a1, q, n):
    conditions(a1, q, n)
    if q == 1:
        return n * a1
    else:
        return a1 * (1 - q ** n) / (1 - q)


if __name__ == '__main__':
    print(a_n(1, 2, 10))
    print(s_n(1, 2, 10))

等差数列

• 公差用d表示
  $n\in N^{\ast }$
• 通项公式：$a_n=a_1+(n+1)\ast d$
• 等差数列求和：$S_n=n\cdot a_1+\frac{n(n-1)}{2}d$

def a_n(a1, d, n):
    assert isinstance(n, int) and n > 0
    return a1 + (n - 1) * d


def s_n(a1, d, n):
    return (a1 + a_n(a1, d, n)) * n / 2


if __name__ == '__main__':
    print(a_n(1, 1, 10))
    print(s_n(1, 1, 10))

斐波那契数列

• 从第3项开始，每一项都等于前两项之和：1，1，2，3，5，8，13，21，34…
• $n\in N^{\ast }$
  $$\begin{gathered} a_1=1 \\ {a}_2=1 \end{gathered}$$
  $当n\ge 3时,a_n=a_{n-1}+a_{n-2}$

def fibonacci(n):
    assert isinstance(n, int) and n > 0
    a = 1
    b = 1
    i = 1
    while i < n:
        a, b = b, a + b
        i += 1
    return a


if __name__ == '__main__':
    for j in range(1, 10):
        print('f(%d)=%d' % (j, fibonacci(j)))

f(1)=1
f(2)=1
f(3)=2
f(4)=3
f(5)=5
f(6)=8
f(7)=13
f(8)=21
f(9)=34