Here you will learn formula to find middle term in binomial expansion with examples.
Letโs begin โย
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 \(({2\over 3}x^2 โ {3\over 2x})^{20}\).
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 = \(T_{11}\)
\(T_{11}\) = \(T_{10 + 1}\) = \(^{20}C_{10}\) \(({2\over 3}x^2)^{20 โ 10}\) \((-{3\over 2x})^{10}\)
= \(^{20}C_{10} x^{10}\)
Example : Find the middle term in the expansion of \((3x โ {x^3\over 6})^7\).
Solution : The given expression is \((3x โ {x^3\over 6})^7\).
Here n = 7, which is an odd number.
So, \(({7 + 1\over 2}\)) th and \(({7 + 1\over 2} + 1)\) th i.e.ย 4th and 5th terms are two middle terms.
Now, \(T_{4}\) = \(T_{3 + 1}\) = \(^{7}C_{3}\) \((3x)^{7 โ 3}\) \((-{x^3\over 6})^{3}\)
\(\implies\) \(T_{4}\) = \((-1)^3\) \(^{7}C_{3}\) \((3x)^{4}\) \(({x^3\over 6})^{3}\)
\(\implies\) \(T_{4}\) = -35 \(\times\) 81 \(x^4\) \(\times\) \(x^9\over 216\) = -\(105x^{13}\over 8\)
and,ย \(T_{5}\) = \(T_{5 + 1}\) = \(^{7}C_{4}\) \((3x)^{7 โ 4}\) \((-{x^3\over 6})^{4}\)
\(\implies\) \(T_{5}\) = \(^{7}C_{4}\) \((3x)^{3}\) \(({x^3\over 6})^{4}\)
\(\implies\) \(T_{5}\)= 35 \(\times\) 27 \(x^3\) \(\times\) \(x^{12}\over 1296\) = \(35x^{15}\over 48\)
Hence, the middle terms are -\(105x^{13}\over 8\) and \(35x^{15}\over 48\).
