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

  • Thread starter mahmudarif
  • Start date
  • #1
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.
 

Answers and Replies

  • #2
606
1
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
 
  • #3
695
2
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.
 
  • #4
that's simple re arrangement of binomial expansions. for general proof, try general form of binomial expansions. you will get it easily.
 
  • #5
984
174
Or else use the "sup" and "sub" tags to create superscripts and subscripts:

[noparse]xy xy[/noparse]

xy xy
 
  • #6
Got the point. Thanks every one...
 

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