# Binomial theorem question

• November 2nd 2008, 07:12 AM
smwatson
Binomial theorem question
If n≥8, and the coefficients of $x^7$ and $x^8$ in the expansion of $(3+x/2)^n$ are equal, what is n?

I equated the coefficents, but end up with n being 8.03 something, which can't be right.

Any help?
• November 2nd 2008, 07:37 AM
Peritus
$
\left( {3 + \frac{x}
{2}} \right)^n = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}c}
n \\
k \\

\end{array} } \right)3^k \left( {\frac{x}
{2}} \right)^{n - k} }
$

now we equate the aforementioned coefficients:

$
\left( {\begin{array}{*{20}c}
n \\
{n - 7} \\

\end{array} } \right)3^{n - 7} \left( {\frac{1}
{2}} \right)^7 = \left( {\begin{array}{*{20}c}
n \\
{n - 8} \\

\end{array} } \right)3^{n - 8} \left( {\frac{1}
{2}} \right)^8
$

$\begin{gathered}
\Leftrightarrow 6\frac{{n!}}
{{7!\left( {n - 7} \right)!}} = \frac{{n!}}
{{8!\left( {n - 8} \right)!}} \hfill \\
\Leftrightarrow 48 = n - 7 \Rightarrow n = 55 \hfill \\
\end{gathered}$