F.4 AM

1a)Prove, by mathematical induction, that
1^2+2^2+......+n^2=(1/6)*n(n+1)(2n+1) for all positive integers n.
Ans:
設P(n) 為命題 , "1^2+2^2+......+n^2=(1/6)*n(n+1)(2n+1)"
當n=1 時 ,
左方=1^2=1
右方=(1/6)(1)(2)(3)=1=左方//
所以P(1)成立.

設對於正整數k , P(k) 成立, 即"1^2+2^2+...+k^2=(1/6)(k)(k+1)(k+2)"

當n=k+1時
左方
=1^2+2^2+....+k^2+(k+1)^2
=(1/6)(k)(k+1)(k+2)+(k+1)^2
=(1/6)(k+1)[k(2k+3)+6(k+1)]
=(1/6)(k+1)(2k^2+3k+6k+6)
=(1/6)(k+1)(k+2)(2k+3)
=右方//

所以P(n) 成立.
根據數學歸納法, 對於正整數n , P(n) 成立.
1b)Using the fomula in (a), find the sum 1x2+2x3+....+n(n+1)
Ans:
1x2+2x3+....+n(n+1)
=1+1+(4+2)+...+n^2+n
=1^2+2^2+...+n^2+1+2+...+n
=(1/6)(n)(n+1)(2n+1)+(1/2)(n)(n+1)
=(1/6)(n)(n+1)[(2n+1)+3]
=(1/3)(n)(n+1)(n+2)//

2a)Prove, by mathematical induction, that (1/1x2x3x4)+(1/2x3x4 x5)+.....+[1/n(n+1)(n+2)(n+3)]=(1/18)-[1/3(n+1)(n+2)(n+3)] for all positive integers n.
Ans:
設P(n) 為命題 , "(1/1x2x3x4)+(1/2x3x4 x5)+.....+[1/n(n+1)(n+2)(n+3)]=(1/18)-[1/3(n+1)(n+2)(n+3)]"
當n=1 時 ,
左方=1/(1x2x3x4)=1/24
右方=1/18-(1/3)(2)(3)(4)=1/24=右方//
所以P(1)成立.

設對於正整數k , P(k) 成立, 即"(1/1x2x3x4)+(1/2x3x4 x5)+.....+[1/k (k +1)(k +2)(k +3)]=(1/18)-[1/3(k +1)(k +2)(k +3)]"
當n=k+1時
左方
=(1/1x2x3x4)+(1/2x3x4 x5)+.....+[1/k (k +1)(k +2)(k +3)}+[1/(k+1)(k+2)(k+3)(k+4)]
=(1/18)-[1/3(k +1)(k +2)(k +3)+1/(k+1)(k+2)(k+3)(k+4)
=(1/18)-[1/3(k)(k+1)(k+2)(k+3)(k+4)](-k-4-k)
=1/18 - 1/(k+1)(k+2)(k+3)(k+4)//
所以P(n) 成立.
根據數學歸納法, 對於正整數n , P(n) 成立.

2b) (1/18)-[1/3(99+1)(99+2)(99+3)] - (1/1x2x3x4)-(1/2x3x4x5)
=171700/3090600-1/3090600-128775/3090600-25755/3090600
=17169/3090600