求多项式的和:1/1 - 1/3 + 1/5 - 1/7 + 1/9 . . . + 1/(2*n-1)的和

Posted 2021-02-19 touch-prc

"""
    求多项式的和：
    1/1 - 1/3 + 1/5 - 1/7 + 1/9 . . .  + 1/(2*n-1)
    n = 1000000
    1）打印出该多项式的和
    2）打印出该和乘以4的值
"""
#方法一
n = int(input(‘请输入一个数（n）：‘))
count = 1 # 计数器
plus = 0 # 存放 1/1+1/5+1/9...+1/(2n-1)
subtract = 0 # 存放 1/3 + 1/7 + 1/11...+1/(2n-2)
sum = 0 # 存放 plus - subtract 的值
while count <= n:
    if count % 2 == 1:
        plus += 1 / (2 * count - 1)
    else:
        subtract += 1 / (2 * count - 1)
    count +=1
sum = plus - subtract
print(‘该多项式的和为：‘,sum)
print(‘sum x 4 =‘,sum*4)

#方法二
num = int(input(‘请输入一个整数：‘))
count = 1 # 计数器
sign = 1 # 表示正号
sum = 0

while count <= num:
    sum += sign * 1/(2 * count -1)
    sign = sign * -1 # 改变正负号
    count += 1
print(‘该多项式的和为：‘,sum)
print(‘sum x 4 =‘,sum * 4)