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Problem 6

1. Problem

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.

平方和与和平方之差

前十个自然数的平方的和是

\[1^2+2^2+...+10^2=385\]

前十个自然数的和的平方是

\[(1+2+...+10)^2=55^2=3025\]

因此,前十个自然数的平方和与和平方之差是 \(3025-385=2640\) 。

求前一百个自然数的平方的与和平方之差。

2. Solution

这一题非常简单,分别算好平方和与和平方然后相减即可:

(defun eu6-sum-of-s (n)
  (let* ((ls (number-sequence 1 n)))
    (cl-reduce '+ (mapcar (lambda (x) (* x x)) ls))))

(defun eu6-squ-of-s (n)
  (expt (cl-reduce '+ (number-sequence 1 n)) 2))

(- (eu6-squ-of-s 100)
   (eu6-sum-of-s 100))
;; 25164150

当然,我们可以推导其数学公式,然后直接算出来:

\[\sum_{i=1}^{n}i^2 = \frac{n(n+1)(2n+1)}{6}\]

\[(\sum_{i=1}^{n}i)^2 = \frac{n^2(n+1)^2}{4}\]

(let ((n 100))
  (- (/ (* n n (1+ n) (1+ n)) 4)
     (/ (* n (1+ n) (1+ (* n 2))) 6)))