# Middle term of a binomial expansion:

#### Finding the middle term of a binomial expansion:

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.