[Euler-Fermat]

ข้อ 16
$1^2 + 2^2 +... + n^2 = \frac{n(n+1)(2n+1)}{6}$
$พิจารณา 1x2 + 2x3 +3X4 +...+(n-1)n$
$a_n = (n-1)n$
$S_n = \sum_{k = 1}^{n} n^2 - n$
= $\frac{n(n+1)(2n+1)}{6} -\frac{3n(n+1)}{6}$
= $\frac{n(n+1)(n-1)}{3}$
จะได้ $\frac{1^2 + 2^2 +... + n^2}{1x2 + 2x3 +3X4 +...+(n-1)n}$
= $\frac{n(n+1)(2n+1)}{n(n+1)(2n-2)}$
= $\frac{2n+1}{2n-2} = \frac{231}{228}$
$จะได้ 2n+1 = 231$
$n = 115$