# Thread: Binomial expansion question

1. ## Binomial expansion question

Hi everybody,
I am having some trouble with these questions and I'm not really sure where to start.

Find the term independant of $\displaystyle x$ in the expansion of:

$\displaystyle (\frac{1}{3x} - \frac{3x^2}{2})^9$

Find the value of $\displaystyle n$ for which the coefficients of $\displaystyle x, x^2$ and $\displaystyle x^3$ in the expansion of $\displaystyle (1 + x)^n$ form three consecutive terms of an arithmetic series.

Thanks all.

2. Originally Posted by mathsnerd1
Hi everybody,
I am having some trouble with these questions and I'm not really sure where to start.

Find the term independant of $\displaystyle x$ in the expansion of:

$\displaystyle (\frac{1}{3x} - \frac{3x^2}{2})^9$
The general term of $\displaystyle (a+ b)^n$ is $\displaystyle \begin{pmatrix}n \\ i\end{pmatrix}a^ib^{n-i}$.
Here, $\displaystyle a= \frac{1}{3x}$, $\displaystyle b= \frac{3x^2}{2}$ and n= 9 so that is $\displaystyle \begin{pmatrix}9 \\ i\end{pmatrix}\frac{1}{3^ix^i}\frac{3^{9-i}x^{2(9-i)}{2^{9-i}}$

Now, for what value of i is $\displaystyle \frac{x^i}{x^{2(9-i)}}= 1$?

Find the value of $\displaystyle n$ for which the coefficients of $\displaystyle x, x^2$ and $\displaystyle x^3$ in the expansion of $\displaystyle (1 + x)^n$ form three consecutive terms of an arithmetic series.
Those coefficents are, of course, $\displaystyle \begin{pmatrix}n \\ i\end{pmatrix}= n$, $\displaystyle \begin{pmatrix}n \\ 2\end{pmatrix}= \frac{n!}{2(n-1)!}$ which must be equal to n+r, and $\displaystyle \begin{pmatrix}n \\ 3\end{pmatrix}= \frac{n!}{3!(n-3)!}$ which must be equal to n+ 2r, for some r, in order to form an arithmetic series. that gives you two equations to solve for n and r.

Thanks all.

3. Originally Posted by HallsofIvy
The general term of $\displaystyle (a+ b)^n$ is $\displaystyle \begin{pmatrix}n \\ i\end{pmatrix}a^ib^{n-i}$.
Here, $\displaystyle a= \frac{1}{3x}$, $\displaystyle b= \frac{3x^2}{2}$ and n= 9 so that is $\displaystyle \begin{pmatrix}9 \\ i\end{pmatrix}\frac{1}{3^ix^i}\frac{3^{9-i}x^{2(9-i)}{2^{9-i}}$

Now, for what value of i is $\displaystyle \frac{x^i}{x^{2(9-i)}}= 1$?

Where did $\displaystyle i$ come from?

And how did you get $\displaystyle \frac{x^i}{x^{2(9-i)}}= 1$?