求1^2 + 2^2 + ... + n^2的值

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首先给出答案, 即 $1^2 + 2^2 + \cdots + n^2 = \frac{1}{6}n(2n+1)(n+1)$. 以下给出证明过程:

$\because (n+1)^3 = n^3 + 3n^2 + 3n + 1$
$\therefore (n+1)^3 - n^3 = 3n^2 + 3n + 1$
同理可得,
$n^3 - (n-1)^3 = 3(n-1)^2 + 3(n - 1) + 1$
$\cdots$
$2^3 - 1^3 = 3\cdot1^2 + 3\cdot 1 + 1$
将以上各式左右分别相加, 可得
$(n+1)^3 - 1 = 3(1^2 + \cdots + n^2) + 3(1+\cdots + n) + n$
$\Longrightarrow n^3 + 3n^2 + 3n = 3(1^2 + \cdots + n^2) + \frac{3}{2}n(n+1) + n$
$\Longrightarrow 6(1^2+\cdots + n^2) = 2n^3 + 3n^2 + n = n(2n+1)(n+1)$
$\Longrightarrow 1^2 + \cdots + n^2 = \frac{1}{6}n(2n+1)(n+1)$