# Find the coefficient of x^{10} in the expansion of (2x + 3)^{13}. You must compute the binomial...

## Question:

Find the coefficient of {eq}x^{10} {/eq} in the expansion of {eq}(2x + 3)^{13}. {/eq} You must compute the binomial coefficient, but you can leave any exponents in exponential form.

## Binomial Expansion or Newton's Binomial:

Binomial coefficients have many applications in statistics and probability calculations.

If we know the exponent of one of the factors in the expansion of a binomial, the other is calculated by the property that the sum of the factor powers of each term is equal to the power of the binomial.

The formula for the power of a binomial is given by

{eq}(x+b)^n = \displaystyle \sum_{k=0}^n \binom{n}{k} x^{n-k}b^k {/eq}

We can observe that the exponents of {eq}x {/eq} are given by the expression

{eq}n-k {/eq}

For {eq}(2x + 3)^{13} {/eq} we have {eq}n=13 {/eq} therefore, the {eq}x^{10} {/eq} exponent is given by

{eq}n-k=10 {/eq}

Solving for {eq}k {/eq} we have

{eq}13-k=10 \to k=3 {/eq}

The coefficient of {eq}x^{10} {/eq} is obtained for {eq}n=13 {/eq} when {eq}k=3 {/eq} i.e.

{eq}\displaystyle \binom{n}{k}= \displaystyle \binom{13}{3} {/eq}

By calculating the binomial coefficient, we have

{eq}\displaystyle \binom{n}{k}= \dfrac{n!}{k!(n-k)!} {/eq} therefore

{eq}\displaystyle \binom{13}{3}= \dfrac{13!}{3!(13-3)!}=\dfrac{13.12.11.10!}{3.2.10!}=\frac{1716}{6}=286 {/eq}

The coefficient of {eq}x^{10} {/eq} is {eq}286 {/eq}

The term containing {eq}x^{10} {/eq} is given by

{eq}286(2x)^{10}(3)^3=286(2)^{10}x^{10}27=7907328x^{10} {/eq} 