Python 程序：計算數列1² + 2² + … + n²

寫一個 Python 程序計算數列 1 +2 +3 +的和……+n 用 For 循環和函數舉例。

級數 1 +2 +3 +的 Python 和的數學公式。+n =(n(n+1)(2n+1))/6

這個 Python 程序要求用戶輸入任意正整數。接下來，Python 程序使用上面的公式找到系列 12+22+32+…+n2的和。

# Python Program to calculate Sum of Series 1²+2²+3²+….+n²

number = int(input("Please Enter any Positive Number  : "))
total = 0

total = (number * (number + 1) * (2 * number + 1)) / 6

print("The Sum of Series upto {0}  = {1}".format(number, total))

系列 1 +2 +3 +的 Python 和……+n 輸出

Please Enter any Positive Number  : 6
The Sum of Series upto 6  = 91.0

Sum =(Number (Number+1)(2 Number+1))/6
Sum =(6 (6+1)(2 6+1))/6 =>(6 7 13)/6
和輸出，Sum = 91

如果你想讓 Python 顯示序列順序 12+22+32+42+52、我們必須在 If Else 的基礎上增加額外的 For Loop

number = int(input("Please Enter any Positive Number  : "))
total = 0

total = (number * (number + 1) * (2 * number + 1)) / 6

for i in range(1, number + 1):
    if(i != number):
        print("%d^2 + " %i, end = ' ')
    else:
        print("{0}^2 = {1}".format(i, total))

Please Enter any Positive Number  : 7
1^2 +  2^2 +  3^2 +  4^2 +  5^2 +  6^2 +  7^2 = 140.0

這個系列 1 +2 +3 +的 Python 和……+n 程序同上。但是在這個 Python 程序中，我們定義了一個函數來放置邏輯。

def sum_of_square_series(number):
    total = 0

    total = (number * (number + 1) * (2 * number + 1)) / 6

    for i in range(1, number + 1):
        if(i != number):
            print("%d^2 + " %i, end = ' ')
        else:
            print("{0}^2 = {1}".format(i, total))

num = int(input("Please Enter any Positive Number  : "))
sum_of_square_series(num)

Please Enter any Positive Number  : 9
1^2 +  2^2 +  3^2 +  4^2 +  5^2 +  6^2 +  7^2 +  8^2 +  9^2 = 285.0

這裡，我們使用 Python 遞歸函數來求數列 1 +2 +3 +的和。+n .

def sum_of_square_series(number):
    if(number == 0):
        return 0
    else:
        return (number * number) + sum_of_square_series(number - 1)

num = int(input("Please Enter any Positive Number  : "))
total = sum_of_square_series(num)

print("The Sum of Series upto {0}  = {1}".format(num, total))