Find the constant term in the expansion of { (x + (\frac{3}{x})^2)^9 } Find the coefficient of...

Question:

Find the constant term in the expansion of {eq}(x + (\frac{3}{x})^2)^9 {/eq}

Find the coefficient of {eq}(x)^4(y)^9 {/eq} in the expansion of {eq}(x + 2y^3)^7 {/eq}

Newton's Binomial:

Newton's binomial can be solved using the formula that groups the terms into combinations,once resolved the respective operations we can obtain not only the constant term of a binomial of any degree but the coefficient of any of the terms that form it.

Answer and Explanation:

a) Let {eq}\left(x + \left(\dfrac{3}{x}\right)^2\right)^9= \left(x + \dfrac{9}{x^2} \right)^9 {/eq}

Using the formula {eq}(a+b)^n= \sum_{k=0}^n \bigl(\begin{smallmatrix} n \\ k \end{smallmatrix}\bigr) a^{n-k}b^k {/eq} with {eq}a= x , b= \dfrac{9}{x^2} {/eq}and {eq}n= 9 {/eq} gives us

{eq}\left(x + \dfrac{3}{x^2}\right)^9= \binom {9}{0} x^9 \left ( \frac{9}{x^2} \right )^0 + \binom {9}{1} x^8 \left ( \frac{9}{x^2} \right ) + \binom {9}{2} x^7 \left ( \frac{9}{x^2} \right )^2 + \binom {9}{3} x^6 \left ( \frac{9}{x^2} \right )^3 + \binom {9}{4} x^5 \left ( \frac{9}{x^2} \right )^4 + \binom {9}{5} x^4 \left ( \frac{9}{x^2} \right )^5 + \binom {9}{6} x^3\left ( \frac{9}{x^2} \right )^6 + \binom {9}{7} x^2 \left ( \frac{9}{x^2} \right )^7 + \binom {9}{8} x \left ( \frac{9}{x^2} \right )^8+ \binom {9}{9} x^0 \left ( \frac{9}{x^2} \right )^9 {/eq}


Solving we have

{eq}\left(x + \dfrac{3}{x^2}\right)^9= 61236 + x^9 + 81x^6 + 2916x^3 + \dfrac{826686}{x^3}+ \dfrac{7440174}{x^6}+ \dfrac{44641044}{x^9}+ \dfrac{172186884}{x^{12}}+ \dfrac{387420489}{x^{15}}+ \dfrac{387420489}{x^{18}} {/eq}

Thus the term constant in the expansion of {eq}\left(x + \left(\dfrac{3}{x}\right)^2\right)^9 {/eq} is {eq}61236 {/eq}


b) Let {eq}(x + 2y^3)^7 {/eq}


We use the same formula as in the previous part with {eq}a= x, b= 2y^3 {/eq} and {eq}n= 7 {/eq}. So we have

{eq}(x + 2y^3)^7 = 128x^7 + 896x^6y^3 + 2688x^5y^6 + (35)(2x)^4 (2)^3y^9 + 4480x^3y^{12}+ 2688x^2 y^{15}+ 896xy^{18} + 128y^{21} {/eq}

{eq}(x + 2y^3)^7 = 128x^7 + 896x^6y^3 + 2688x^5y^6 + 4480 x^4 y^9 + 4480x^3y^{12}+ 2688x^2 y^{15}+ 896xy^{18} + 128y^{21}. {/eq}


Thus the coefficient of {eq}(x)^4(y)^9 {/eq} in the expansion of {eq}(x + 2y^3)^7 {/eq} is {eq}4480. {/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
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