[s3想的][Maths]連老師都prove不到的問題 值得一看

orb

我s3的[=.=]

Please prove the phenomenon
3^2+4^2=5^2              3^2=9=4+5
5^2+12^2=13^2            5^2=25=12+13
7^2+24^2=25^2            7^2=49=24+25
︴
99^2+4900^2=4901^2       99^2=9801=4900+4901


Answer:

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[ 本帖最後由 ＥnTaＳIa 於 17/5/2007 00:16 編輯 ]

kazuhiko

3^2+4^2=5^2
3^2 = 5^2 - 4^2 = (5-4)(5+4) = 9

In general,
Let a^2 + b^2 = c^2 and c-b=1
a^2 = c^2 - b^2 = (c-b)(c+b) = b+c

Therefore, a^2 + b^2 = c^2  is the same as a^2 = b +c, provided that c-b=1

STEVE

原帖由 ＥnTaＳIa 於 16/5/2007 22:54 發表

                               
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我s3的[=.=]

Please prove the phenomenon
3^2+4^2=5^2              3^2=9=4+5
5^2+12^2=13^2            5^2=25=12+13
7^2+24^2=25^2            7^2=49=24+25
︴
99^2+4900^2+901^2       99^2=9801 ...

其實, 兩個相鄰的平方數之差一定是一個奇數, 但這個奇數是平方數的話
那麼, 我們知道
b=a-1
a^2-b^2=(a+b)(a-b)=c^2
而且a-b=1
所以我們有a^2-b^2=a+b=c^2 and b=a-1

亂來的

Dragonite

我無聊地兜左勁大個圈最後用MI黎prove
不過我無視去到99為止
當佢去到無限
(因為事實上個statement去到無限大都岩)

純粹無聊得濟玩野- -
唔好插我唔該...

我一開波既idea係
只要prove到5+13+25+...+4+12+24+...
               =3^2+5^2+7^2+.....就ok

consider the series 4+12+24+...
It can be written as 4*1+(4*1+4*2)+(4*1+4*2+4*3)+...
                          =4(1+1+2+1+2+3+...+1+......)

Consider the sequence 1+2+3+.....+n
Let n be the no.of the terms

General term (An) = 1+(n-1)1=n-------------(^)    <By An=a+(n-1)d

So,
  4+12+24+..
=4(1+1+2+1+2+3+...+1+2+3......n)             (by (^), general term (An)=n)
  5+13+25+...
=4+12+24+...+n
=4(1+1+2+1+2+3+...+1+2+3......n)+n         (since there are n terms in the series)

Then
  4+12+24+...+5+13+25+...
=4(1+1+2+1+2+3+...+1+2+3......49)+4(1+1+2+1+2+3+...+1+2+3......n)+n
=2*4(1+(1+2)+(1+2+3)+...+(1+2+3......n))+n
=8(1+(1+2)+(1+2+3)+...+(1+2+3......n))+n

We are going to use MI to prove for the truth of the statement.

  8(1+(1+2)+(1+2+3)+...+(1+2+3+...+n))+n
=3^2+(3+2)^2+(3+2+2)^2+...(3+2n-2)^2                                 <BY An=a+(n-1)d, in the sequence 3,5,7,...n , An=(3+(n-1)2)=3+2n-2

Put n=1
LHS=8(1)+1=9
RHS=3^2=9
The1st equ is true

Assume that the k th equ is true.
8(1+(1+2)+(1+2+3)+...+(1+2+3+...+k))+k=3^2+(3+2)^2+(3+2+2)^2+... (3+2k-2)^2
To show it gives (k+1)th equ
  8(1+(1+2)+...+k(k+1)/2+(k+1)(k+1+1)/2)+k+1)
=3^2+(3+2)^2+(3+2+2)^2...+(3+2k-2)^2+(3+2(k+1)-2)^2---------(*)
Add 8(k+1)(k+1+1)/2+1 both sides
8(1+(1+2)+...+k(k+1)/2)+k+1
=3^2+(3+2)^2+(3+2+2)^2+...(3+2k-2)^2+8(k+1)(k+1+1)/2+1
=3^2+(3+2)^2+(3+2+2)^2+...(3+2k-2)^2+4(k^2+3k+2)+1
=3^2+(3+2)^2+(3+2+2)^2+...(3+2k-2)^2+4k^2+12k+9
=3^2+(3+2)^2+(3+2+2)^2+...(3+2k-2)^2+(2k+3)^2
=3^2+(3+2)^2+(3+2+2)^2+...(3+2k-2)^2+(3+2(k+1)-2)^2
=(*)

By MI,the equ is true for n=1,2,3......

[ 本帖最後由 Dragonite 於 4/7/2007 15:40 編輯 ]