Question

How do you find the 5th term in the expansion of the binomial ${{\left( 5a+6b \right)}^{5}}$?

Answer 1

The term of the form ${{\left( a+b \right)}^{n}}$ is called a binomial term. In the expansion of this binomial, there are total n+1 terms. The (r+1)th term of the expansion of the binomial expansion is $^{n}{{C}_{r}}{{a}^{r}}{{b}^{n-r}}$. We can find the binomial term by substituting the values of a, b, and n in this general term. It should be noted that here $n$ is a positive integer.

Complete step by step solution:
We are asked to expand the binomial term ${{\left( 5a+6b \right)}^{5}}$. Comparing with the general binomial term ${{\left( a+b \right)}^{n}}$, we get $a=5a,b=6b\And n=5$. We know the general form of (r+1)th term of the binomial series is $^{n}{{C}_{r}}{{a}^{r}}{{b}^{n-r}}$. As we want to find the 5th term, we have $r+1=5$. From this, we can find the value of r as 4. We can find the required binomial term by substituting the value of variables in the general form, as follows
$^{n}{{C}_{r}}{{a}^{r}}{{b}^{n-r}}$
${{\Rightarrow }^{5}}{{C}_{4}}{{\left( 5a \right)}^{4}}{{\left( 6b \right)}^{5-4}}$
We know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, using this to simplify the above series we get
$^{5}{{C}_{4}}=\dfrac{5!}{4!\left( 5-4 \right)!}=5$
Substituting the values of the above coefficients, we get
$\Rightarrow 5{{\left( 5a \right)}^{4}}\left( 6b \right)$
Simplifying the above term, we get

& \Rightarrow 5\times 625\times 6{{a}^{4}}b=\text{18750}{{a}^{4}}b \\\ & \Rightarrow \text{18750}{{a}^{4}}b \\\ \end{aligned}$$ **Note:** We can use more special binomial expansions to expand the series. If one of the terms inside the bracket is 1. then, we can use the expansion of $${{\left( 1+x \right)}^{n}}$$ whose general form of expansion is $$\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{x}^{r}}}$$. For these types of problems, these expansions are very important and should be remembered. We can use these expansions only when $$n$$ is a positive integer. For cases when the $$n$$ is a non-positive integer, we need to use different types of expansions.