我想問幾條附加數的題目(20點)

Explanation as follows:

圖片參考：http://i238.photobucket.com/albums/ff245/chocolate328154/Maths306.jpg?t=1216097889


2008-07-15 14:04:21 補充：
注意: Assumption嗰步係要指明 [ Assume P ( k ) is true for SOME positive integers k ]；
有d書係寫[ Assume P is true for n = k ], 個n = k同樣表達咗SOME嘅意思
但下面嗰位[Assume P(k) is true,k is a positive integer], 係冇指明到SOME positive integers k

2008-07-15 14:07:04 補充：
lee個純粹係小問題一個, 但唔排除要求嚴格嘅老師會扣掉部分分數= =

2008-07-16 12:20:00 補充：
注意: 003並非完全使用附加數的方法。Comparing the coefficients等是純數常用的方法, 在附加數的課程上是無須用到。

1. Let 1/[(2n-1)(2n+1)]＝a/(2n-1)＋b/(2n+1)
＝＞　　1＝a(2n+1)＋b(2n-1)
＝＞　　1＝2(a＋b)n＋(a－b)

Comparing the coefficients of n: 2(a＋b)＝0　＝＞　a＋b＝0
Comparing the constant terms: a－b＝1

Hence, a＝0.5 and b＝－0.5

Therefore

1/[(2n-1)(2n+1)]＝0.5/(2n-1)－0.5/(2n+1)

and

1/(1×3)＋1/(3×5)＋1/(5×7)＋．．．＋1/[(2n－1)(2n＋1)]＝(0.5/1－0.5/3)＋(0.5/3－0.5/5)＋(0.5/5－0.5/7)＋．．．＋[0.5/(2n-1)－0.5/(2n+1)]
＝0.5/1＋(－0.5/3＋0.5/3)＋(－0.5/5＋0.5/5)＋．．．＋[－0.5/(2n-1)＋0.5/(2n-1)]－0.5/(2n+1)
＝0.5－0.5/(2n+1)
＝n/(2n+1)


2. 9^n－1＝9^n＋[9^(n－1)－9^(n－1)]＋．．．＋(9²－9²)＋(9－9)－1
＝[9^n＋9^(n－1)＋．．．＋9²＋9]－[9^(n－1)＋．．．＋9²＋9＋1]
＝9[9^(n－1)＋．．．＋9²＋9＋1]－[9^(n－1)＋．．．＋9²＋9＋1]
＝(9－1)[9^(n－1)＋．．．＋9²＋9＋1]
＝8[9^(n－1)＋．．．＋9²＋9＋1]


3. 2^(2n)－1＝4^n－1
＝4^n＋[4^(n－1)－4^(n－1)]＋．．．＋(4²－4²)＋(4－4)－1
＝[4^n＋4^(n－1)＋．．．＋4²＋4]－[4^(n－1)＋．．．＋4²＋4＋1]
＝4[4^(n－1)＋．．．＋4²＋4＋1]－[4^(n－1)＋．．．＋4²＋4＋1]
＝(4－1)[4^(n－1)＋．．．＋4²＋4＋1]
＝3[4^(n－1)＋．．．＋4²＋4＋1]

圖片參考：http://www.photo-host.org/img/041263screenhunter_02_jul._15_13.17.gif

2.Let P(n) be the propostion
“9n-1is divisible by 8”
For n=1,
91-1=8 is divisible by 8.
∴P(1) is also true
Assume P(k) is true,k is a positive integer.
i.e. “9k-1=8N,N is an integer ”
For n=k+1,
9k+1-1=9(9k)-1
=9(8N+1)-1
=9(8N)+8
=8(9N+1) which is divisibly by 8.
∴P(k+1) is also true
By principle of mathematical induction ,P(n) is true for all positive intergers n.

3.Let P(n) be the propostion
“22n-1is divisible by 3.”
For n=1,
22-1=3 is divisible by 3.
∴P(1) is also true
Assume P(k) is true,k is a positive integer.
i.e. “22k-1=3N,N is an integer ”
For n=k+1,
22k+2-1=4(22k)-1
=4(3N+1)-1
=3(4N)+3
=3(4N+1) which is divisibly by 3.
∴P(k+1) is also true
By principle of mathematical induction ,P(n) is true for all positive intergers n.



2008-07-20 21:29:10 補充：
我答慢了!所以最佳解答應給Gabriella Montez !!!