Project Euler

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# [[Problem 6]]

## Sum Square Difference

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

$$1^2 + 2^2 + ... + 10^2 = 385.$$

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

$$(1 + 2 + ... + 10)^2 = 55^2 = 3025.$$

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.

## 二乗和の差

最初の10個の自然数について, その二乗の和は,

$$1^2 + 2^2 + ... + 10^2 = 385.$$

最初の10個の自然数について, その和の二乗は,

$$(1 + 2 + ... + 10)^2 = 55^2 = 3025.$$

これらの数の差は 3025 - 385 = 2640 となる.

同様にして, 最初の100個の自然数について二乗の和と和の二乗の差を求めよ.

```scheme

#!r6rs

#!chezscheme

(import (chezscheme))

(define square-num-list

  (map

    (lambda (n) (* n n))

    (iota 101)))

(define square-sum

  (apply + square-num-list))

(define sum-square

  (expt (apply + (iota 101))

        2))

(define answer-6

  (- sum-square square-sum))

(printf "6: ~D~%" answer-6)

```