# Middle term in the expansion $\rm (x-\frac2x)^{16}$ is

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Middle term in the expansion $\rm (x-\frac2x)^{16}$ is
1. $\rm ^{16}C_9 \times 2^8$
2. $\rm ^{16}C_9 \times 2^9$
3. $\rm ^{16}C_8 \times 2^8$
4. $\rm ^{16}C_8 \times 2^9$

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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

∴ $\rm T_{9}=T_{8+1}$ $\rm =(-1)^8\times^{16}C_8\times x^{16-16} \times 2^8$

⇒ $\rm ^{16}C_8 \times 2^8$