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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.

Answer and Explanation:


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}


Learn more about this topic:

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How to Use the Binomial Theorem to Expand a Binomial

from Algebra II Textbook

Chapter 21 / Lesson 16
10K

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