Question

Find the middle term in the expansion of ${\left( {2x + 3y} \right)^8}$.

Answer 1

Here we will first find the total number of terms in the expansion of the given terms. Then we will use the general term of the binomial expansion to get the middle term in the expansion. The binomial theorem is defined as a method to expand the expression with some finite power raised to it.

Complete step by step solution:
Here we need to find the middle terms in the expansion of the given algebraic expression.
The given binomial is ${\left( {2x + 3y} \right)^8}$.
We know from the binomial theorem that if ${\left( {a + b} \right)^n}$ is the binomial then there will be total $n + 1$ terms in the binomial expansion.
Therefore, the total number of terms in the given binomial $= 8 + 1 = 9$.
We know from the binomial theorem that if $n$ is even then middle will be equal to ${\left( {\dfrac{n}{2} + 1} \right)^{th}}$.
On substituting the value of $n$ here, we get
Middle Term $= {\left( {\dfrac{8}{2} + 1} \right)^{th}}$
$\Rightarrow$ Middle Term $= {\left( {4 + 1} \right)^{th}} = {5^{th}}$ term
The general term find the ${r^{th}}$ term of the binomial expansion is given as ${T_r} = {}^n{C_r} \cdot {a^{n - r + 1}} \cdot {b^{r - 1}}$.
Now, we will use this general term to find the middle term i.e. ${5^{th}}$ term of the binomial expansion.
${T_5} = {}^8{C_{5 - 1}} \cdot {\left( {2x} \right)^{8 - 5 + 1}} \cdot {\left( {3y} \right)^{5 - 1}}$
On further simplifying the terms, we get
$\Rightarrow {T_5} = {}^8{C_4} \cdot {2^4} \cdot {x^4} \cdot {3^4} \cdot {y^4}$
Now, we will apply the exponents on the bases.
$\Rightarrow {T_5} = {}^8{C_4} \cdot 16 \cdot {x^4} \cdot 81 \cdot {y^4}$
On multiplying the terms, we get
$\Rightarrow {T_5} = {}^8{C_4} \cdot 1296 \cdot {x^4}{y^4}$

Using this formula of combination ${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}}$, we get
$\Rightarrow {T_5} = \dfrac{{8!}}{{\left( {8 - 4} \right)! \times 4!}} \cdot 1296 \cdot {x^4}{y^4}$
On further simplifying the terms, we get
$\Rightarrow {T_5} = \dfrac{{8!}}{{4! \times 4!}} \cdot 1296 \cdot {x^4}{y^4}$
Now, we will find the value of the factorials.
$\Rightarrow {T_5} = \dfrac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4! \times 4 \times 3 \times 2 \times 1}} \cdot 1296 \cdot {x^4}{y^4}$
On further simplifying the terms, we get
$\Rightarrow {T_5} = 90720{x^4}{y^4}$

This is the required middle term in the expansion of ${\left( {2x + 3y} \right)^8}$.

Note:
In the above solution, we have used the term ‘binomial theorem’. The binomial theorem of expansion is generally used in algebra, probability, etc. Here, we have also used the combination formula to simplify the expression. The combination is a way or method of selecting an element from a set of elements such that the order of selection does not matter. It is different from permutation and therefore we must be careful while applying the formula. Permutation is a way or method of arranging an element from a set of elements such that the order of arrangement matters.