 Prove, by mathematical induction, that
(n+1)(n+2)(n+3)...(2n)=(2^n)×1×3×5×...×(2n-1)
for all positive integers n.

solution: (by asdfzxc806)
when n=1
LHS=2
RHS=(2)(1)=2
P(1) is true
when n=k+1
LHS=(k+2)(k+3)(k+4)...(2k)(2k+1)(2k+2)
=(2^k)×1×3×5×...×(2k-1)(2k+1)(2k+2)/(k+1)
=2^(k+1)x1x3x5x...x(2k+1)
RHS=2^(k+1)x1x3x5x...x(2k+1)
P(k+1) is true

Given that a and b are positive integers. By mathematical induction, show that
(a+b)^(2n+1)-a^(2n+1)-b^(2n+1) is divisible by ab(a+b) for all positive integers n.

If it looks too difficult, you may first try the special case a=2, b=3.
52n+1-22n+1-32n+1 is divisible by 30 for all positive integers na

when n=1
P(n)=(a+b)^3-a^3-b^3
=3a^2b+3ab^2
=3ab(a+b)   which is divisible by ab(a+b)
Assume that P(k) is true
i.e. (a+b)^(2k+1)-a^(2k+1)-b^(2k+1)=Mab(a+b)     where M is an integer

[ 本帖最後由 bc123456 於 2010-5-3 12:26 PM 編輯 ]

L.H.S. = (2^k)(2k-1)(2k+2)
R.H.S. = (2^(k+1))(2k+1)

##### 引用:

Prove, by mathematical induction, that
(n+1)(n+2)(n+3)...(2n)=(2^n)×1×3×5×...×(2n-1)
for all positive integers n.
when n=1
LHS=2
RHS=(2)(1)=2
P(1) is true
when n=k+1
LHS=(k+2)(k+3)(k+4)...(2k)(2k+1)(2k+2)
=(2^k)×1×3×5×...×(2k-1)(2k+1)(2k+2)/(k+1)
=2^(k+1)x1x3x5x...x(2k+1)
RHS=2^(k+1)x1x3x5x...x(2k+1)
P(k+1) is true

when n=1, RHS=(2^1)×(2*1-1)=2
##### 引用:

LHS=(k+2)(k+3)(k+4)...(2k)(2k+1)(2k+2)
=(2^k)×1×3×5×...×(2k-1)(2k+1)(2k+2)/(k+1)

How can you prove???

Try this one. Given that a and b are positive integers. By mathematical induction, show that
(a+b)[suptag]2n+1[/suptag]-a[suptag]2n+1[/suptag]-b[suptag]2n+1[/suptag] is divisible by ab(a+b) for all positive integers n.

If it looks too difficult, you may first try the special case a=2, b=3.
5[suptag]2n+1[/suptag]-2[suptag]2n+1[/suptag]-3[suptag]2n+1[/suptag] is divisible by 30 for all positive integers n
(a question that appeared in this forum a few times)

##### 引用:

when n=1
LHS=2
RHS=(2)(1)=2
P(1) is true
when n=k+1
LHS=(k+2)(k+3)(k+4)...(2k)(2k+1)(2k+2)
=(2^k)×1×3×5×...×(2k-1)(2k+1)(2k+2)/(k+1)
=2^(k+1)x1x3x5x...x(2k+1)
RHS=2^(k+1) ...

(2^k)x1x3x5x...(2k-1)(2k+1)(2k+2)/(k+1)
=(2^k)x1x3x5...(2k-1)(2k+1)2(k+1)/(k+1)

=(2^k+1)x1x3x5x...(2k-1)(2k+1) [ 本帖最後由 felix139139139 於 2010-5-3 11:40 AM 編輯 ]

LHS=(k+2)(k+3)(k+4)...(2k)(2k+1)(2k+2)
=(k+1)(k+2)(k+3)(k+4)...(2k)(2k+1)(2k+2)/(k+1)
=(2^k)×1×3×5×...×(2k-1)(2k+1)(2k+2)/(k+1)         (assumption,(k+1)(k+2)(k+3)(k+4)...(2k)=2[suptag]k[/suptag]×1×3×5×...×(2k-1) )

[ 本帖最後由 lopo42 於 2010-5-3 11:15 AM 編輯 ]

##### 引用:

(2^k)x1x3x5x...(2k-1)(2k+1)(2k+2)/(k+1)
=(2^k)x1x3x5...(2k-1)(2k+1)2(k+1)/(k+1)

=(2^k+1)x1x3x5x...(2k-1)(2k+1) 1x3x5x7x9x...x21=1x3x5x7x9x...x19x21

1x3x5x...(2n+1)=1x3x5x...(2n-5)(2n-3)(2n-1)(2n+1)

##### 引用:

Try this one. Given that a and b are positive integers. By mathematical induction, show that
(a+b)2n+1-a2n+1-b2n+1 is divisible by ab(a+b) for all positive integers n.

If it looks too diff ...

##### 引用:

1x3x5x7x9x...x21=1x3x5x7x9x...x19x21

1x3x5x...(2n+1)=1x3x5x...(2n-5)(2n-3)(2n-1)(2n+1)

push...睇下我做得啱唔啱

push...睇下我做得飱唔飱

##### 引用:

[ 本帖最後由 bc123456 於 2010-5-3 01:33 PM 編輯 ]

##### 引用:

Try this one. Given that a and b are positive integers. By mathematical induction, show that
(a+b)2n+1-a2n+1-b2n+1 is divisible by ab(a+b) for all positive integers n.

If it looks too diff ...
alternative method

##### 引用:

alternative method
It works.

By the way, this can be solved much easier by modular arithmetic.

Using mathematical induction is really really really tedious.

##### 引用:

bc123456 , 加油

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