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.

By using middle terms in binomial expansion formula,

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


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