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Can anyone please help me to generalize and prove this if it is valid?

 
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Apr22-12, 03:00 PM   #1
 

Can anyone please help me to generalize and prove this if it is valid?


X^4-(X^3)y+(X^2)(y^2)-x(y^3)+y^4= (x+y)^4-5xy(x+y)^2+5(xy)^2

x^6-(x^5)y+(X^4)(y^2)-(X^3)(y^3)+(X^2)(y^4)-x(y^5)+y^6=(x+y)^6-7xy(x+y)^4 +14((xy)^2)(x+y)^2 - 7(xy)^3

....................
If n is an odd prime then prove,

x^n-1 - X^(n-2).y+..........-x.y^(n-2)+y^(n-1) = (x+y)^n-1 - nxy(x+y)^(n-3) +..........(-1)^((n-1)/2) . n .(xy) ^ ((n-1)/2)


Thank you very much in advance for your assistance.
 
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Apr22-12, 05:18 PM   #2
 
Quote by mahmudarif View Post
X^4-(X^3)y+(X^2)(y^2)-x(y^3)+y^4= (x+y)^4-5xy(x+y)^2+5(xy)^2

x^6-(x^5)y+(X^4)(y^2)-(X^3)(y^3)+(X^2)(y^4)-x(y^5)+y^6=(x+y)^6-7xy(x+y)^4 +14((xy)^2)(x+y)^2 - 7(xy)^3

....................
If n is an odd prime then prove,

x^n-1 - X^(n-2).y+..........-x.y^(n-2)+y^(n-1) = (x+y)^n-1 - nxy(x+y)^(n-3) +..........(-1)^((n-1)/2) . n .(xy) ^ ((n-1)/2)


Thank you very much in advance for your assistance.


Your post is very difficult to read in ASCII...I even didn't try. My advice: learn how to post here in LaTeX or else

attach some document, preferably PDF, where mathematical stuff appears decently.

DonAntonio
 
Apr23-12, 02:58 AM   #3
 
I think this has been posted before recently but, for the love of me, I can't find the original post.

And I believe the suggestion was to learn about the binomial theorem; then try to expand (x+y)^4 and (x+y)^6.
 
Jun27-12, 08:26 AM   #4
 

Can anyone please help me to generalize and prove this if it is valid?


that's simple re arrangement of binomial expansions. for general proof, try general form of binomial expansions. you will get it easily.
 
Jul3-12, 09:43 AM   #5
 
Or else use the "sup" and "sub" tags to create superscripts and subscripts:

x[sup]y[/sup] x[sub]y[/sub]

xy xy
 
Aug13-12, 07:51 PM   #6
 
Got the point. Thanks every one...
 
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