# Homework Help: Binomial Expansion

1. Nov 22, 2011

### thornluke

The coefficient of x in the expansion of [x+(1/ax^2)]^7 is 7/3. Find the possible values of a.

1. Rewrite (x + 1/(ax^2))^7 = x^(-14) (x^3 + 1/a)^7.
So, we need to find the coefficient of x^15 from (x^3 + 1/a)^7.

2. Using the Binomial Theorem, we have
(x^3 + 1/a)^7 = Σ(k = 0 to 7) C(7, k) (x^3)^(7 - k) (1/a)^k.
...................= Σ(k = 0 to 7) C(7, k) x^(21 - 3k) (1/a)^k.

3. So, we need 21 - 3k = 15 ==> k = 2.

Thus, we have (1/a)^2 = 7/3
==> a = ±√(3/7).

The problem is, I do not understand the steps. Help please?

2. Nov 22, 2011

### grzz

First $\frac{1}{x^{2}}$ is taken out as a common factor. Note that this factor is to the power of 7.

3. Nov 22, 2011

### thornluke

Why do we need to find the coefficient of x^15 from (x^3 + 1/a)^7?

4. Nov 22, 2011

### Ray Vickson

How else could you get x^1 when multiplying x^(-14) and x^n?

RGV

5. Nov 23, 2011

### thornluke

Oh right.. Thank you! :)

6. Nov 23, 2011

### HallsofIvy

I wouldn't have factored out the $x^{14}$. Each term of $(u+ v)^n$ is of the form
$$\begin{pmatrix}n \\ i\end{pmatrix}u^iv^{n-i}$$

Here, u= x and $a= 1/ax^2= (1/a)x^{-2}$ so that formula would be
$$\begin{pmatrix}7 \\ i\end{pmatrix}a^{i- 7}(x^i)(x^{2i- 14})$$
and we want the power of x, i+ 2i- 14= 1. That gives 3i= 15 or i= 5.

It looks to me like you have forgotten the binomial coefficient. I get an integer value for a.

7. Nov 23, 2011

### thornluke

I don't understand how $$\begin{pmatrix}7 \\ i\end{pmatrix}a^{i- 7}$$ can come from $$u^iv^{n-i}$$

If it's n-i shouldn't it be a7-i and not ai-7?