Ta thấy: $1^{2}+2^{2}+...+n^{2}$ là đa thức bậc 3
Giả sử: P(x)= $ax^{3}+bx^{2}+cx+d$ ($a\neq 0$)
Ta cần tìm a, b, c, d
P(1)= a+b+c+d = 1$^{2}$
P(2)= 8a+4b+2c+d= $1^{2}+2^{2}$
P(3)= 27a+9b+3c+d= $1^{2}+2^{2}+3^{2}$
P(4)= 64a +16b+4c+d= $1^{2}+2^{2}+3^{2}+4^{2}$
$\Rightarrow$ $a=\frac{1}{3}$; $b=\frac{1}{2}$; $c=\frac{1}{6}$; d=0
$\Rightarrow$ P(x)= $1^{2}+2^{2}+...+n^{2}$= $\frac{1}{3}x^{3}+\frac{1}{2}x^{2}+\frac{1}{6}x$= $\frac{n(n+1)(2n+1)}{6}$
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