# Middle Term in Binomial Expansion

Here you will learn formula to find middle term in binomial expansion with examples.

Let’s begin –

## Middle Term in Binomial Expansion

Since the binomial expansion of $$(x + a)^n$$ contains (n + 1) terms. Therefore,

(1) If n is even, then $${n\over 2} + 1$$ th term is the middle term.

(2) If n is odd, then $${n + 1\over 2}$$ th and $${n + 3\over 2}$$ th terms are the two middle terms.

Example : Find the middle term in the expansion of $$({2\over 3}x^2 – {3\over 2x})^{20}$$.

Solution : Here, n = 20, which is an even number.

So, $${20\over 2} + 1$$th term i.e. 11th term is the middle term.

Hence, the middle term = $$T_{11}$$

$$T_{11}$$ = $$T_{10 + 1}$$ = $$^{20}C_{10}$$ $$({2\over 3}x^2)^{20 – 10}$$ $$(-{3\over 2x})^{10}$$

= $$^{20}C_{10} x^{10}$$

Example : Find the middle term in the expansion of $$(3x – {x^3\over 6})^7$$.

Solution : The given expression is $$(3x – {x^3\over 6})^7$$.

Here n = 7, which is an odd number.

So, $$({7 + 1\over 2}$$) th and $$({7 + 1\over 2} + 1)$$ th i.e.  4th and 5th terms are two middle terms.

Now, $$T_{4}$$ = $$T_{3 + 1}$$ = $$^{7}C_{3}$$ $$(3x)^{7 – 3}$$ $$(-{x^3\over 6})^{3}$$

$$\implies$$ $$T_{4}$$ = $$(-1)^3$$ $$^{7}C_{3}$$ $$(3x)^{4}$$ $$({x^3\over 6})^{3}$$

$$\implies$$ $$T_{4}$$ = -35 $$\times$$ 81 $$x^4$$ $$\times$$ $$x^9\over 216$$ = -$$105x^{13}\over 8$$

and,  $$T_{5}$$ = $$T_{5 + 1}$$ = $$^{7}C_{4}$$ $$(3x)^{7 – 4}$$ $$(-{x^3\over 6})^{4}$$

$$\implies$$ $$T_{5}$$ = $$^{7}C_{4}$$ $$(3x)^{3}$$ $$({x^3\over 6})^{4}$$

$$\implies$$ $$T_{5}$$= 35 $$\times$$ 27 $$x^3$$ $$\times$$ $$x^{12}\over 1296$$ = $$35x^{15}\over 48$$

Hence, the middle terms are -$$105x^{13}\over 8$$ and $$35x^{15}\over 48$$.