nth term of binomial expansion with negative exonent

**rcs** · September 20th 2012, 08:31 AM

can anybody share on me how to find the nth term of a binomial expansion / theorem with the negative exponet

(x+y)^-2,, find the 9th term.. how is it possible?

thanks

**johnsomeone** · September 20th 2012, 10:15 PM

I don't think what you're looking for exists.

There is this: For $|a|<1, \frac{1}{1-a} = 1 + a + a^2 + a^3 +...$ .

So if $|x+y| < 1$ , then $\frac{1}{(x+y)^2} = \frac{1}{1 - ( 1 - (x+y)^2)} = 1 + a + a^2 + a^3 +...$ , where $a = 1 - (x+y)^2$ (Note $0 < 1 - (x+y)^2 < 1$ ).

You could expand that out, giving:

$\frac{1}{(x+y)^2} = 1 + (1 - (x+y)^2) + (1 - (x+y)^2)^2 + (1 - (x+y)^2)^3 +...$

$= 1 + (1 - (x+y)^2) + (1 -2(x+y)^2 + (x+y)^4) + ...$

$= 1 + (1 - (x^2+2xy+y^2) + (1 -2(x^2+2xy+y^2) + (x^4+4x^3y+6x^2y^2+4xy^3+y^4)) + ...$

And then collect all like terms - if it even converges term-wise when you try that (I don't know), much less converges to the original value. So, you could maybe give this approach a whirl - no guarantees from me that it gets you anywhere though. And it puts you in a world of series and convergence - a bit different than the nice formula for the binomial coefficients. I don't think the thing you want exists.

**a tutor** · September 21st 2012, 12:47 AM

Maybe $(x+y)^{-2}=\left(x\left(1+\frac{y}{x}\right)\right)^{-2}=x^{-2}\left(1+\frac{y}{x}\right)^{-2}$ and then expand to get $x^{-2}\left(1-\frac{2y}{x}+\frac{3y^2}{x^2}-\frac{4y^3}{x^3}...$

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