# How do you find the coefficient of x^6 in the expansion of (2x+3)^10?

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6
Jan 4, 2016

Calculate the binomial coefficient and appropriate powers of $2$ and $3$ to find that the required coefficient is:

$1088640$

#### Explanation:

${\left(2 x + 3\right)}^{10} = {\sum}_{k = 0}^{10} \left(\begin{matrix}10 \\ k\end{matrix}\right) {2}^{10 - k} {3}^{k} {x}^{10 - k}$

The term in ${x}^{6}$ is the one for $k = 4$, so has coefficient:

$\left(\begin{matrix}10 \\ 4\end{matrix}\right) {2}^{6} \cdot {3}^{4}$

=(10!)/(4! 6!)*64*81

$= \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \cdot 64 \cdot 81$

$= 210 \cdot 64 \cdot 81 = 1088640$

Instead of calculating $\left(\begin{matrix}10 \\ 4\end{matrix}\right)$, you can pick it out from the appropriate row of Pascal's triangle...

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