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。
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)))