Question

Find the middle term in the expansion of \(\rm (4x-\frac{x^3}{6})^6\)
1. \(\rm \frac{27}{80}x\)
2. -\(\rm \frac{27}{16} x^{12}\)
3. \(\rm \frac{8}{27} x^{13}\)
4. -\(\rm \frac{160}{27} x^{12}\)

Answer 1

Correct Answer - Option 4 : -\(\rm \frac{160}{27} x^{12}\)

Concept:

The general term in the expansion of (a - b)ⁿ is given by, \(\rm T_{r+1}=^{n}C_r\times a^{n-r}\times b^r\)

• When n is even, then the middle term = \(\rm (\frac n 2+1)\)th term.
• When n is odd, then the middle term = \(\rm \frac 12(n+1)\)th term and \(\rm \frac 12(n+3)\)th term.

 

Calculation:

GIven expansion = \(\rm (4x-\frac{x^3}{6})^6\)

The general term in the given expansion is given by,
 \(\rm T_{r+1}=(-1)^r\times ^{6}C_r\times(4x)^{6-r}\times(\frac{x^3}{6})^r\)

 

Now, n is even, so middle term = \(\rm (\frac n 2+1)\)th term = 3 + 1 = 4th term 

\(\rm T_4=T_{3+1}\) = \(\rm (-1)^3\times ^6C_3\times(4x)^3 \ ({x^3\over 6})^{3}\)

= - \(\rm \frac{6\times5\times4 }{3\times 2\times1}\times 4^{3}\times \frac{1}{6^3}\times x^{12}\)

= - \(\rm \frac{160}{27} \cdot x^{12}\)

Hence, option (4) is correct.