關於A.Maths (20點)

Mathematical Induction 1x2+2x4+3x5+…+n(n+1)=1/3 n(n+1)(n+2)
得一個方法就係用數學歸納法= = 無公式用既.....我做一次啦~

設s(n)為命題
1x2+2x4+3x5+…+n(n+1)=1/3 n(n+1)(n+2)

考慮s(1)
左=1(1+1)=2
右=1/3 1(1+1)(1+2)=2
所以s(1)成立

假設s(k)成立
考慮s(k+1)
左
=1x2+2x4+3x5+…+k(k+1)+(k+1)[(k+1)+1]
=1/3 k(k+1)(k+2) +(k+1)[(k+1)+1]
=1/3 k(k+1)(k+2) +(k+1)(k+2)
=1/3 (k+1)(k+2) (k+3).................................抽個1/3 (k+1)(k+2)出泥
=1/3 (k+1)[(k+1)+1][(k+1)+2]
=右

所以s(k+1)成立
根據數學歸納法的原理,對於所有自然數n , s(n)都成立


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let P(n) be the proposition
"1x2+2x4+3x5+…+n(n+1)=1/3 n(n+1)(n+2) ----(*)"
when n = 1, LHS = 1(1+1) = 2
RHS = 1/3 1(1+1)(1+2)
=2
=LHS
∴P(1) is true.

assume P(k) is true for some positive integer k.
i.e. 1x2+2x4+3x5+…+k(k+1)=1/3 k(k+1)(k+2)
Then 1x2+2x4+3x5+…+k(k+1) + (k+1)[(k+1)+1] <---呢個係下個term
=1/3 k(k+1)(k+2) + (k+1)[(k+1)+1]
=1/3 [k(k+1)(k+2) + 3(k+1)(k+2)]<-----通分母
=1/3 (3+k)(k+1)(k+2) <-------抽common factor
=1/3 (k+1)[(k+1)+1][(k+1)+2]

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

As follow AS~~~

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Let P(n) be the proposition

when n=1
LHS:1(1+1)=2
RHS:1/3*1*(1+1)(1+2)=2 =LHS
so LHS=RHS
SO P(1) is true

assume P(k) is true for some postive integer k,
1x2+2x4+3x5+…+k(k+1)= 1/3k(k+1)(k+2)

when n=k+1

LHS:1x2+2x4+3x5+…+1/3k(k+1)(k+2)+k+1(k+2)
=1/3(k+1)(k+2)[k+3]
=1/3(k+1)(k+2)(k+3)

RHS:1/3(k+1)(k+1+1)(k+2+1)
=1/3(k+1)(k+2)(k+3)

SO LHS=RHS
SO P(k+1) is true

By the principle of MI ,P(n) is true for all positive integer n.

M.I.唔係用黎淨係用LHS去搵RHS出黎，而且M.I.唔係公式。
Mathematical Induction
1x2+2x4+3x5+…+n(n+1)=1/3 n(n+1)(n+2)
For all positive integer n.
Prove: Let the statment be S(n).
For n = 1
LHS = 1*2
= 2
RHS = 1/3 n(n+1)(n+2)
= 1/3 (2)(3)
= 2
= LHS
So, S(1) is true.
Let S(k) is true.
i.e. 1*2+2*4+3*5+…+k(k+1)=1/3 k(k+1)(k+2)
For n = k+1
LHS = 1*2+2*4+3*5+…+k(k+1)+(k+1)(k+2)
= 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)
= RHS
So, S(k+1) is also true.
By M.I., S(n) is true for all +ve integer n.