數學歸立法

Let P(n) be the statement that

12 + 22 + … + n2 = 1/6 n(n+1)(2n+1) for all positive integers n

When n = 1

L.H.S. = 12 = 1

R.H.S. = 1/6 (1)(1+1)[2(1)+1] = 1

∴ P(1) is true.

Assume P(k) is true for some positive integers k

12 + 22 + … + k2 = 1/6 k(k+1)(2k+1)

When n = k+1

L.H.S. = 12 + 22 + … + k2 + (k+1)2

= 1/6 k(k+1)(2k+1) + (k+1)2

= 1/6 (k+1)[k(2k+1) + 6(k+1)]

= 1/6 (k+1)(2k2 + 7k + 6)

= 1/6 (k+1)(k+2)(2k+3)

= 1/6 (k+1)[(k+1)+1][2(k+1)+1]

= R.H.S.

∴ By the principle of mathematical induction, P(n) is true for all positive integers n.

設S(n)為命題
"1²+ 2²+...+n²= 1/6 n(n+1)(2n+1)"
當n=1時,
左方=1²
=1
右方=1/6 (1)(1+1)(2+1)
=1
∴左方=右方
所以,S(1)成立。
假設對於所以正整數k,命題S(k)成立,
即1²+ 2²+...+k²= 1/6 k(k+1)(2k+1)。
當n=k+1時,
左方=1²+ 2²+...+k²+(k+1)²
=1/6 k(k+1)(2k+1)+(k+1)²
=1/6(k+1)[k(2k+1)+6(k+1)]
=1/6(k+1)(2k²+k+6k+6)
=1/6(k+1)(2k²+7k+6)
=1/6(k+1)(k+2)(2k+3)
=右方
所以,S(k+1)成立。
根據數學歸納法,對於所以正整數n,命題S(n)都成立。