Question

Find the middle term (s) in the expansion of the following ${\left( {\dfrac{a}{b} + \dfrac{a}{b}} \right)^6}$.

Answer 1

Hint: Before attempting this question prior knowledge of binomial expansion, formulas belonging to the concept of binomial expansion are must, use the given details will help you to approach towards the solution to the problem.

Complete step-by-step solution -
According to the given information we have binomial ${\left( {\dfrac{a}{b} + \dfrac{a}{b}} \right)^6}$
So we know that by the binomial theorem in the given binomial it will have n+1 terms here n is 6
So the total numbers of terms in the given binomial expansion will be 6 + 1 = 7
By the binomial theorem if in a binomial number n is an even number then the middle term will be ${\left( {\dfrac{n}{2} + 1} \right)^{th}}$ term
So the middle term of binomial number ${\left( {\dfrac{a}{b} + \dfrac{a}{b}} \right)^6}$ will be ${\left( {\dfrac{n}{2} + 1} \right)^{th}}$ term of the expansion
Middle term of expansion = ${\left( {\dfrac{6}{2} + 1} \right)^{th}}$= $4^{\text{th}}$ term of the expansion will be the middle term of the expansion
To find the $\text{(r+1)}^{\text{th}}$ term of expansion equation is given as ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}$
So let’s find out the $4^{\text{th}}$ term of the expansion
Substituting the given values in the above equation
${T_{3 + 1}}{ = ^6}{C_3}{\left( {\dfrac{a}{b}} \right)^{6 - 3}}{\left( {\dfrac{a}{b}} \right)^3}$
So we know that $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Therefore ${T_{3 + 1}} = \dfrac{{6!}}{{3!\left( {6 - 3} \right)!}}{\left( {\dfrac{a}{b}} \right)^{6 - 3}}{\left( {\dfrac{a}{b}} \right)^3}$
\Rightarrow $$${T_{3 + 1}} = \dfrac{{6!}}{{3!3!}}{\left( {\dfrac{a}{b}} \right)^{6 - 3}}{\left( {\dfrac{a}{b}} \right)^3}$$ We know that n! =n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ..... \times 2 \times 1 Therefore $${T_{3 + 1}} = \dfrac{{\left( {6 \times 5 \times 4 \times 3 \times 2 \times 1} \right)}}{{\left( {3 \times 2 \times 1} \right)\left( {3 \times 2 \times 1} \right)}}{\left( {\dfrac{a}{b}} \right)^3}{\left( {\dfrac{a}{b}} \right)^3}$$ \Rightarrow {T_{3 + 1}} = 20{\left( {\dfrac{a}{b}} \right)^3}{\left( {\dfrac{a}{b}} \right)^3}$$ $ \Rightarrow {T_{3 + 1}} = 20{\left( {\dfrac{a}{b}} \right)^6} Hence the $4^{th}$ term (middle term) of the given binomial expansion is20{\left( {\dfrac{a}{b}} \right)^6}$$.

Note: In the above solution we used a term binomial theorem according to which it is a method to expand the expression with some finite power raised to it. This method of expansion is generally used in algebra, probability, etc. So if for any expression general expansion is given by $\sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}}.{y^r}{ + ^n}{C_r}{x^{n - r}}.{y^r} + .....{ + ^n}{C_{n - 1}}x.{y^{n - 1}}{ + ^n}{C_n}.{y^n}}$ here $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.