Question

How do you find the ${{8}^{th}}$ term in the expansion of the binomial ${{\left( 4x+3y \right)}^{9}}$?

Answer 1

The expression of the type ${{\left( a+b \right)}^{n}}$ is called a binomial expression. The expansion of this expression in summation form is written as $\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}}$. There are total n+1 terms in the expansion. The general term is written as ${{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$. Here $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.

Complete answer:
We are given the binomial ${{\left( 4x+3y \right)}^{9}}$, and we have to find the ${{8}^{th}}$ term in the expansion. We know that for a general binomial of form ${{\left( a+b \right)}^{n}}$ the general formula for the ${{\left( r+1 \right)}^{th}}$ term is ${{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$. Here we have, a = 4x, b = 3y, and n = 9.
We have to find the ${{8}^{th}}$, which means
$\Rightarrow r+1=8$
Subtracting 1 from both sides of the equation, we get

& \Rightarrow r+1-1=8-1 \\\ & \therefore r=7 \\\ \end{aligned}$$ Now we have the value of r for finding the $${{8}^{th}}$$ term. We can find the term by substituting the value of r in the standard form of the term. By doing this we get, $$\begin{aligned} & \Rightarrow {{T}_{7+1}}{{=}^{9}}{{C}_{7}}{{\left( 4x \right)}^{9-7}}{{\left( 3y \right)}^{7}} \\\ & \Rightarrow {{T}_{8}}{{=}^{9}}{{C}_{7}}{{\left( 4x \right)}^{2}}{{\left( 3y \right)}^{7}} \\\ \end{aligned}$$ We know that the value of $${{4}^{2}}$$ is 16, and the value of $${{3}^{7}}$$ can be calculated which equals 2187. We know that $$^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$$ we have n = 9, and r = 7. Substituting these values, we get $${{\Rightarrow }^{9}}{{C}_{7}}=\dfrac{9!}{7!\left( 9-7 \right)!}=\dfrac{9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{\left( 7\times 6\times 5\times 4\times 3\times 2\times 1 \right)\left( 2\times 1 \right)}$$ By canceling out the common factors form numerators and denominator, the above expression can be written as $$\Rightarrow \dfrac{9\times 8}{2\times 1}=\dfrac{72}{2}=36$$ Substituting this value in the expression for the eighth term, we get $$\begin{aligned} & \Rightarrow {{T}_{8}}{{=}^{9}}{{C}_{7}}{{\left( 4x \right)}^{2}}{{\left( 3y \right)}^{7}}=36\times {{\left( 4x \right)}^{2}}{{\left( 3y \right)}^{7}} \\\ & \Rightarrow 36\times \left( 16{{x}^{2}} \right)\left( 2187{{y}^{7}} \right) \\\ & \Rightarrow 36\times 16\times 2187\left( {{x}^{2}} \right)\left( {{y}^{7}} \right) \\\ & \Rightarrow 1259712{{x}^{2}}{{y}^{7}} \\\ \end{aligned}$$ Hence the $${{8}^{th}}$$ term of the given binomial is $$1259712{{x}^{2}}{{y}^{7}}$$. **Note:** To solve these types of problems one should know the expansion of different types of binomial expression and their general term formula. Also, one should know how to find the value of $$^{n}{{C}_{r}}$$.