# Binomial expansion question

• Mar 6th 2011, 12:35 PM
mathsnerd1
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 $x$ in the expansion of:

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

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

Thanks all.
• Mar 6th 2011, 12:43 PM
HallsofIvy
Quote:

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 $x$ in the expansion of:

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

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

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

Quote:

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

Quote:

Thanks all.
• Mar 6th 2011, 01:01 PM
mathsnerd1
Quote:

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

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

Where did $i$ come from? :confused:

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