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標題: 數論證明1(已解決) [列印本頁]

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作者: 傲月光希    時間: 07-9-5 11:38
標題: 數論證明1(已解決)
一篇回覆請回答一題，過程必須清楚，正確無誤者給予3枚聲望作獎勵

1.分兩小題，請利用第一數歸法，須完全證明便有獎勵
(a)1+3+5+...+(2n-1)=n^2，對所有正整數n。
(b)1*2+2*3+3*4+...+n(n+1)=n(n+1)(n+2)/3，對所有正整數n。

2.請利用第二數歸法證明(a^n)-1=(a-1)[a^(n-1)+a^(n-2)+...+1]，對所有正整數n。
[提示：已知a^(n+1)-1=(a+1)(a^n-1)-a(a^(n-1)-1)]

3.用數學歸納法證明1(1!)+2(2!)+...+n(n!)=(n+1)!-1，對所有正整數n。

[ 本文最後由 傲月光希 於 07-9-22 01:10 PM 編輯 ]

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作者: aeoexe    時間: 07-9-5 16:51
1.分兩小題，請利用第一數歸法，須完全證明便有獎勵
(a)1+3+5+...+(2n+1)=n^2，對所有正整數n。<- 懷疑出錯,如果n=1, L.H.S.=4,R.H.S.=1
(b)1*2+2*3+3*4+...+n(n+1)=n(n+1)(n+2)/3，對所有正整數n。


(a)Let S(n) be the statement,'1+3+5+...+(2n-1)=n^2'for all natural no. n

When n=1 L.H.S.=1 R.H.S.=1^2=1 Therefore, S(1) is true.

Assume S(k) is true,i.e. 1+3+5+...+(2k-1)=k^2

When n=k+1,

L.H.S.=1+3+5+...+2k-1+2k+1

=k^2+2k+1

=(k+1)^2

=R.H.S.

Therefore,S(k+1) is true
By the principal of Mathematical Induction, S(n) is true for all positive integers n

(b) Let S(n) be the statement,'1*2+2*3+3*4+...+n(n+1)=n(n+1)(n+2)/3'for all natural no. n

When n=1 L.H.S.=1*2=2 R.H.S.=1*2*3/3=2 Therefore, S(1) is true.

Assume S(k) is true,i.e. 1*2+2*3+...+k(k+1)=k(k+1)(k+2)/3

When n=k+1,

L.H.S.=1*2+2*3+....+k(k+1)+(k+1)(k+2)

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

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

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

=(k+1)[(k+1)+1][(k+1)+2]/3

=R.H.S.

Therefore,S(k+1) is true.
By the principal of Mathematical Induction, S(n) is true for all positive integers n

[ 本文最後由 aeoexe 於 07-9-5 05:32 PM 編輯 ]

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作者: aeoexe    時間: 07-9-5 16:56
3.用數學歸納法證明1(1!)+2(2!)+...+n(n!)=(n+1)!-1，對所有正整數n。

Let S(n) be the statement,'1(1!)+2(2!)+...+n(n!)=(n+1)!-1'for all natural no. n

When n=1, L.H.S.=1 R.H.S.=2!-1=1 Therefore, S(1) is true.

Assume S(k) is true,i.e. 1(1!)+2(2!)+...+k(k!)=(k+1)!-1

When n=k+1,

L.H.S.=1(1!)+2(2!)+...+k(k!)+(k+1)(k+1)!

=(k+1)!+(k+1)(k+1)!-1

=(k+2)(k+1)!-1

=(k+2)!-1

=[(k+1)+1]!-1

=R.H.S.

Therefore,S(k+1) is true.
By the principal of Mathematical Induction, S(n) is true for all positive integers n

[ 本文最後由 aeoexe 於 07-9-5 05:32 PM 編輯 ]

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作者: aeoexe    時間: 07-9-5 17:31
2.請利用第二數歸法證明(a^n)-1=(a-1)[a^(n-1)+a^(n-2)+...+1]，對所有正整數n。
[提示：已知a^(n+1)-1=(a+1)(a^n-1)-a(a^(n-1)-1)]

Let S(n) be the statement,'(a^n)-1=(a-1)[a^(n-1)+a^(n-2)+...+1]'for all natural no. n

(1) When n=1, L.H.S.=a-1 R.H.S.=(a-1)(1) S(1)is true.

(2)Assume S(k) is true,When n=k+1,

L.H.S=a^(k+1)-1

=(a+1)(a^k-1)-a(a^(k-1)-1)

=(a+1)(a^k-1)-a^k-a

=(a-1)(a+1)[a^(k-1)+a^(k-2)+...+1]-a^k+a

=(a-1)[a^k+2a^(k-1)+2a^(k-2)+...+1]-a^k+a

=(a-1)[a^k+a^(k-1)+a^(k-2)+...+1]+(a-1)[a^(k-1)+a^(k-2)+...+a+1-1]-a^k+a

=(a-1)[a^k+a^(k-1)+a^(k-2)+...+1]+a^k-1-a+1-a^k+a

=(a-1)[a^k+a^(k-1)+a^(k-2)+...+1]

=R.H.S.

Therefore,S(k+1) is true.
By (1)(2) and the principal of Mathematical Induction, S(n) is true for all positive integers n

[ 本文最後由 aeoexe 於 07-9-5 05:39 PM 編輯 ]

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