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 \(35x^{15}\over 48\).

Leave a Comment

Your email address will not be published.