Correct Answer - Option 3 :
\(\rm ^{16}C_8 \times 2^8\)
Concept:
The general term in the expansion of (a - b)n is given by, \(\rm T_{r+1}=(-1)^r \times ^nC_ra^{n-r}b^r\)
- 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:
The general term in the expansion of \(\rm (x-\frac2x)^{16}\)is given by,
\(\rm T_{r+1}=(-1)^r\times^{16}C_r\times x^{16-r}\times(\frac2x)^r\)
\(\rm ⇒ (-1)^r\times ^{16}C_r\times x^{16-2r}\times 2^{r}\)
Here, n is 16 (i.e., even), so there will be one middle terms
\(\rm (\frac n2+1)\)th term = (8 + 1)th term = 9th term
∴ \(\rm T_{9}=T_{8+1}\) \(\rm =(-1)^8\times^{16}C_8\times x^{16-16} \times 2^8\)
⇒ \(\rm ^{16}C_8 \times 2^8\)