Question

Find the middle terms in the expansion of ${(3x-\frac{x^3}{6})}^7$

Or

Find the fourth term from the end in the expansion of ${(\frac{3}{x^2}-\frac{x^3}{6})}^7$

Answer 1

The given expression is ${(3x-\frac{x^3}{6})}^7$

Here, n=7 which is an odd number.

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

$Now,T_4{=}^7{C}_3(3x{)}^4{\left(\frac{-x^3}{6}\right)}^3$

$=-\frac{7!}{3!4!}\times (3{)}^4.(x{)}^4\times \frac{\left(x^9\right)}{6^3}$

$=-35\times 81\times \frac{x^{13}}{216}=-\frac{105}{8}\times x^{13}$

and $T_5{=}^7{C}_4(3x{)}^3{(-\frac{x^3}{6})}^4$

$=\frac{7!}{4!3!}\times 3^3\times x^3\times \frac{x^{12}}{6^4}$

$=353\times 27\times \frac{x^{15}}{1296}=\frac{35}{48}\times x^{15}$

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

Or

We have, ${(\frac{3}{x^2}-\frac{x^3}{6})}^7$

Clearly, the given expansion contains 8 terms.

So, 4th term from the end $=(8-4+1)$th =5th term from the beginning

$\therefore$ Fifth term.$T_5{=}^7{C}_4{\left(\frac{3}{x^2}\right)}^{7-4}{(-\frac{x^3}{6})}^4$

${=}^7{C}_4\frac{3^3}{x^6}\times \frac{x^{12}}{6^4}=\frac{7!}{4!3!}\times \frac{27\times x^6}{1296}=\frac{35}{48}{x}^6$

Hence , fourth term from the end in the expansion of ${(\frac{3}{x^2}-\frac{x^3}{6})}^{7}is\frac{35}{48}{x}^6.$