Case(1): If "n" is an even number then the number of terms of the binomial expansion will be (n + 1), which definitely is an odd number. Hence middle term will be \(\left(\frac{n}{2} + 1\right)^{th}\) term. $$ k = \frac{n}{2} + 1 $$ We do not need to use any different formula for finding the middle term of the binomial expansion. The same formula of \(T_k\) will be used for the middle term also. $$ T_k = \binom{n}{k - 1} \ a^{(n - k + 1)} \ b^{(k - 1)} $$
Example: Find the middle term of a binomial \((2x + 3y)^8\)?
Solution: Given, \(n = 8\), \(a = 2x\), \(b = 3y\)
Here n = 8, which is an even number, hence $$ k = \frac{n}{2} + 1 $$ $$ k = \frac{8}{2} + 1 $$ $$ k = 5^{th} \ term $$
Now putting all the values in the formula of \(T_k\). $$ T_k = \binom{n}{k - 1} \ a^{(n - k + 1)} \ b^{(k - 1)} $$ $$ T_5 = \binom{8}{5 - 1} \ (2x)^{(8 - 5 + 1)} \ (3y)^{(5 - 1)} $$ $$ T_5 = \binom{8}{4} \ (2x)^4 \ (3y)^4 $$ $$ T_5 = \frac{8!}{(8 - 4)! \ 4!} \ (2x)^4 \ (3y)^4 $$ $$ T_5 = \frac{8!}{4! \ 4!} \ (2x)^4 \ (3y)^4 $$ $$ T_5 = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4! \ 4!} \ (16x^4) \ (81y^4) $$ $$ T_5 = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \ (16x^4) \ (81y^4) $$ $$ T_5 = 90,720 \ x^4 \ y^4 $$
Case(2): If "n" is an odd number then the number of terms of the binomial expansion will be (n + 1), which definitely is an even number. Hence in this case there will be two middle terms \((\frac{n + 1}{2})\) and \((\frac{n + 3}{2})\).
Example: Find the middle term of a binomial \((2 + 3y)^9\)?
Solution: Given, \(n = 9\), \(a = 2\), \(b = 3y\)
Here n = 9, which is an odd number, hence there will be two middle terms of the binomial expansion. $$ k_1 = \frac{n + 1}{2} $$ $$ k_1 = \frac{9 + 1}{2} $$ $$ k_1 = 5^{th} \ term $$ and $$ k_2 = \frac{n + 3}{2} $$ $$ k_2 = \frac{9 + 3}{2} $$ $$ k_2 = 6^{th} \ term $$
Let's take \(k = 5\) first and put all the values in the formula of \(T_k\). $$ T_k = \binom{n}{k - 1} \ a^{(n - k + 1)} \ b^{(k - 1)} $$ $$ T_5 = \binom{9}{5 - 1} \ (2)^{(9 - 5 + 1)} \ (3y)^{(5 - 1)} $$ $$ T_5 = \binom{9}{4} \ (2)^5 \ (3y)^4 $$ $$ T_5 = \frac{9!}{(9 - 4)! \ 4!} \ (2)^5 \ (3y)^4 $$ $$ T_5 = \frac{9!}{5! \ 4!} \ (32) \ (81y^4) $$ $$ T_5 = \frac{9 \times 8 \times 7 \times 6 \times 5!}{5! \ 4!} \ (32) \ (81y^4) $$ $$ T_5 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \ (32) \ (81y^4) $$ $$ T_5 = 326,592 \ y^4 $$
Now we have to find out the second middle term, which is \(6^{th}\) term. Take \(k = 6\) and find the second middle term yourself.