Correct Answer - Option 3 : Both 1 and 2
Concept:
The middle term of the expansion of (x + a)n depends upon the value of n.
(i) When n is even the total number of terms in the expansion of (x + a)n is n+1(odd).So, there is only one middle term, i.e \(\rm\left ( \frac{n+1}{2} \right )\)th term is the middle term.
(ii) When n is odd The total number of terms in the expansion of (x + a)n is n+1(even).So, there is two middle term, i.e., \(\rm\left ( \frac{n+1}{2} \right )\)th and \(\rm\left ( \frac{n+3}{2} \right )\)th term are two middle terms.
Calculation:
The number of terms in the expansion of \(\rm \left ( 4x-\frac{x^3}{2} \right )^9\) is 10(even). so there are two middle terms. i.e. \(\rm\left ( \frac{9+1}{2} \right )\)th and \(\rm\left ( \frac{9+3}{2} \right )\) th terms. they are given by T5 and T6.
Tr+1 = nCrxn-rar
Required term
T5 = T4+1 = 9C4(4x)5(-x3/2)4
\(\rm T_{5}= {}^9C_{4}4^5x^5\left ( \frac{1}{2^4} \right )x^{12}\)
\(\rm T_{5}= {}^9C_{4}4^3x^{17}\)
and T6 = T5+1 = 9C5(4x)4(-x3/2)5
\(\rm T_{6}= -{}^9C_{5}4^4x^4\left ( \frac{1}{2^5} \right )x^{15}\)
\(\rm T_{6}=- {}^9C_{5}\left ( \frac{2^8}{2^5} \right )x^{19}\)
\(\rm T_{6}=- {}^9C_{5}\ 2^3\ x^{19}\)
