數學歸納法

(i)1+2+3+...+n = (1/2)n( n+1)
(1/2)(1)(1+1) = 1
假設 1+2+3+...+k = (1/2)(k)(k+1)
考慮 1+2+3+...+k + (k+1)
= (1/2)(k)(k+1) + (k+1)
= (1/2)(k+1) [ k + 2 ]
= (1/2)(k+1) [ (k + 1) + 1 ]
當 n = k 情況啱時, n = k+1 情況就啱
其他野自己寫埋佢

(ii)1x2+2x3+3x4+...+ n(n+1) = (1/3)n(n+1)(n+2 )
(1/3)(1)(1+1)(1+2 ) = 2 = (1)(1+1)
假設 1x2+2x3+3x4+...+k (k+1) = (1/3)(k)(k+1)(k+2 )
考慮 1x2+2x3+3x4+...+k (k+1) + (k+1)(k+1+1)
= (1/3)(k)(k+1)(k+2 ) + (k+1)(k+2)
= (1/3)(k+1)(k+2 ) [ k + 3 ]
= (1/3)(k+1) [(k+1) + 1] [ (k + 1) + 2 ]
當 n = k 情況啱時, n = k+1 情況就啱
其他野自己寫埋佢

(b)利用(a)公式,求1²+2²+3²+...+n²公式
1x2+2x3+3x4+...+ n(n+1)
= 1x1 + 1 +2x2 + 2 +3x3 + 3 +...+ ( n)(n)+ n
= (1²+2²+3²+...+n²)+ (1+2+3+...+n)
所以 (1²+2²+3²+...+n²)
= 1x2+2x3+3x4+...+ n(n+1) - (1+2+3+...+n)
= (1/3)n(n+1)(n+2) - (1/2)n( n+1)
= (1/6)n(n+1) [ 2(n+2) - 3 ]
= (1/6)n(n+1) [ 2n+1 ]