Binomial theorem question

**smwatson** · November 2nd 2008, 07:12 AM

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?

**Peritus** · November 2nd 2008, 07:37 AM

$ \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 $

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

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