# Thread: Binomial Expansions with negative and fractional Expontents

1. ## Binomial Expansions with negative and fractional Expontents

I know, I know two threads in 5 minutes, but I can't get this one!

Binomial expansions are all fine and dandy, but when the exponent is a negative, a fraction, or both I am a little stumped.

Write the fourth term in the binomial expansion for:
(a - b)^-(1/5)

and I don't even know where to begin!

2. Originally Posted by Crell
I know, I know two threads in 5 minutes, but I can't get this one!

Binomial expansions are all fine and dandy, but when the exponent is a negative, a fraction, or both I am a little stumped.

Write the fourth term in the binomial expansion for:
(a - b)^-(1/5)

and I don't even know where to begin!
Are you familiar with the binomial theorem?

Binomial theorem - Wikipedia, the free encyclopedia

3. Originally Posted by adkinsjr
Are you familiar with the binomial theorem?

Binomial theorem - Wikipedia, the free encyclopedia

Yes, but I don't know how to apply it with negative, and fractional exponents.

Binomial expansions are all fine and dandy, but when the exponent is a negative, a fraction, or both I am a little stumped.

4. Originally Posted by Crell
Yes, but I don't know how to apply it with negative, and fractional exponents.
Ok, I think it's the same basic idea, you would just have an infinite number of terms. It's not easy to do.

http://www.mathhelpforum.com/math-he...exponents.html

5. Originally Posted by Crell
I know, I know two threads in 5 minutes, but I can't get this one!

Binomial expansions are all fine and dandy, but when the exponent is a negative, a fraction, or both I am a little stumped.

Write the fourth term in the binomial expansion for:
(a - b)^-(1/5)

and I don't even know where to begin!
If $\displaystyle a \neq 0$ I would divide by $\displaystyle a$

$\displaystyle \left[a\left(1-\frac{b}{a} \right)\right]^{-\frac{1}{5}} = a^{-\frac{1}{5}} \times \left(1-\frac{b}{a}\right)^{-\frac{1}{5}}$

The first part is easy enough and the second part can be expressed by manipulating

$\displaystyle (1+x)^n = 1+nx+\frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ... + \frac{n!}{n!}x^n$

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