F4 maths ( MI ) 廿分哦 第二題...

a)
when n=1 ,
1*7=1/6*1*(1+1)(2+19)=7
Assume for some positive integer k
i.e. 1x7+2x8+3x9+...+k(k+6)=1/6k(k+1)(2k+19)
For n=k+1
1x7+2x8+3x9+...+k(k+6)+(k+1)(k+7)
=1/6k(k+1)(2k+19)+(k+1)(k+7) (by assumption)
=1/6(k+1)[k(2k+19)+6(k+7)]
=1/6(k+1)(2k^2+25k+42)
=1/6(k+1)(k+2)(2k+21)
=1/6(k+1)(k+2)[2(k+1)+19]
By M.I. , the statement is true for all positive integers n
b)
1x2+2x3+3x4+...+n(n+1)
=1x(7-5)+2x(8-5)+3x(9-5)+...+n[(n+6)-5]
=(1x7-5)+(2x8-2x5)+(3x9-3x5)+...+n(n+6)-5n
=1x7+2x8+3x9+...+n(n+6)-(5+2x5+3x5+...+5n)
=1x7+2x8+3x9+...+n(n+6)-5(1+2+3+...+n)
=1/6n(n+1)(2n+19)-5n(n+1)/2
=n^3/3+n^2+2n/3
=1/3n(n+1)(n+2)


2009-07-31 00:05:22 補充：
一般做MI時都會刪去"Let P(n) be the statement "...." "呢句唔寫的
同埋記住MI係對於非負整數時先用到,對實數係用唔到
同埋假設ASSUMPTION個度要睇番題目對什麼數去證明,如正整數/非負整數,要寫清楚
b part都係基本的延伸題,想法儘量寫成a part的樣式
for all positive integers n= ∀n∈ℕ
∀=for all
∈=belongs to
ℕ=natural number set

2009-07-31 00:09:09 補充：
btw, 你應該係預備新高中學制的M2課程嗎?

a).
Let P(n) be the statement "1x7+2x8+3x9+...+n(n+6)=1/6n(n+1)(2n+19)"
for all positive integers n.
when n=1 ,
L.H.S.=1x7=7
R.H.S.=1/6(1)(1+1)(2x1+19)=7
so P(1) is true
Assume P(k) is true for some positive integers k,
i.e.1x7+2x8+3x9+...+k(k+6)=1/6k(k+1)(2k+19)
when n=k+1,
L.H.S.=1x7+2x8+3x9+...+k(k+6)+(k+1)(k+1+6)
　　　=1/6k(k+1)(2k+19)+(k+1)(k+7)　　　[by the assumption]
　　　=1/6(k+1)(2k^2+19k+6k+42)
　　　=1/6(k+1)(2k^2+25k+42)
　　　=1/6(k+1)(k+2)(2k+21)
R.H.S.=1/6(k+1)(k+1+1)[2(k+1)+19]
　　　=1/6(k+1)(k+2)(2k+21)
so P(k+1) is true
By the principle of MI ,P(n) is ture for all positive integers n

b).1x2+2x3+3x4+...+n(n+1)
=1(7-5)+2(8-5)+3(9-5)+...+n(n+6-5)
=[1x7+2x8+3x9+...+n(n+6)]-5(1+2+3+...+n)
=1/6n(n+1)(2n+19)-5n(n+1)/2　　[by the result of (a)]
=n(n+1)(2n+19-5x3)/6
=1/3n(n+1)(n+2)


我唔係用英文讀,d文法可能有d錯

係