問題

問題文 (意訳)

100までの自然数の和の2乗と自然数の2乗の和の差を求めよ

原文

The sum of the squares of the first ten natural numbers is,

\begin{align}
1^2 + 2^2 + ... + 10^2 = 385
\end{align}

The square of the sum of the first ten natural numbers is,

\begin{align}
(1 + 2 + ... + 10)^2 = 55^2 = 3025
\end{align}

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is
3025 - 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

答え

25164150

解答の方針

1個目の解答方針

\begin{align}
1^2 + 2^2 + ... + n^2 &= \frac{1}{6}n(n+1)(2n+1)\ \\
(1 + 2 + ... + n)^2 &= \bigl(\frac{1}{2}n(n+1)\bigr)^2\ \\

\end{align}

を使って
以下のように式変形し、手で計算する。

\begin{align}

(1 + 2 + ... + n)^2  - (1^2 + 2^2 + ... + n^2) \\
\end{align}

\begin{align}
&=(\frac{1}{2}n(n+1))^2\ -\frac{1}{6}n(n+1)(2n+1)\\
&= n(n+1)(\frac{1}{4}n(n+1) - \frac{1}{6}(2n+1)) \\
&= \frac{1}{12}(n-1)n(n+1)(3n+2) \\
\end{align}

100の場合だと 99×100×101×302÷12　となって手で計算できる

2個目の解答方針

単純に計算する

Pythonのコード

Python Ploblem6

def sum_of_squares(limit: int) -> int:
    sum: int = 0
    for i in range(limit + 1):
       sum +=  i**2
    return sum
def sum_squared(limit: int) -> int:
    sum: int = 0
    for i in range(limit + 1):
        sum += i
    return sum**2

print(sum_squared(100)-sum_of_squares(100))

Juliaのコード

Julia Ploblem6

function sum_of_squares(limit::Int64)::Int64
    sum::Int64 = 0
    for i in 1:limit
       sum +=  i^2
    end
    return sum
end

function sum_squared(limit::Int64)::Int64
    sum::Int64 = 0
    for i in 1:limit
        sum += i
    end
    return sum^2
end

println(sum_squared(100)-sum_of_squares(100))