The general term of (1+2x)12 is

(1+2x)12=n(n−1)(n−2)⋯n−(r−1)r!.xr=1212−112−2⋯(12−(r−1))(−1)r−1(2x)rr!=12−32−52⋯(2n−32)(−1)r−1.2r.xrr!=1.3.5⋯(2n−3)2r.(−1)r−12r.xrr!=1.3.5⋯(2n−3).(−1)r−1.xrr!

The general term of (1+2x)12 is

(1+2x)12=n(n−1)(n−2)⋯n−(r−1)r!.xr=1212−112−2⋯(12−(r−1))(−1)r−1(2x)rr!=12−32−52⋯(2n−32)(−1)r−1.2r.xrr!=1.3.5⋯(2n−3)2r.(−1)r−12r.xrr!=1.3.5⋯(2n−3).(−1)r−1.xrr!

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The general term of ${(1 + 2 x)}^{\frac{1}{2}}$ is

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$\frac{1.3.5 \dots (2 r - 3) {(- 1)}^{r - 1} x^{r}}{r!}$

$\frac{1.3.5 \dots (2 r - 1) {(- 1)}^{r - 1} x^{r}}{r!}$

$\frac{1.3.5 \dots (2 r - 5) {(- 1)}^{r - 1} x^{r}}{r!}$

$\frac{1.3.5 \dots (2 r - 1) {(- 1)}^{r - 1} x^{r}}{(r + 1)!}$

answer is A.

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Detailed Solution

$\begin{array}{l} {(1 + 2 x)}^{\frac{1}{2}} \\ = \frac{n (n - 1) (n - 2) \dots n - (r - 1)}{r!} . x^{r} \\ = \frac{\frac{1}{2} (\frac{1}{2} - 1) (\frac{1}{2} - 2) \dots (\frac{1}{2} - (r - 1)) {(- 1)}^{r - 1} {(2 x)}^{r}}{r!} \\ = \frac{(\frac{1}{2}) (- \frac{3}{2}) (\frac{- 5}{2}) \dots (\frac{2 n - 3}{2}) {(- 1)}^{r - 1} {.2}^{r} . x^{r}}{r!} \\ = \frac{\frac{1.3.5 \dots (2 n - 3)}{2^{r}} . {(- 1)}^{r - 1} 2^{r} . x^{r}}{r!} \\ = \frac{1.3.5 \dots (2 n - 3) . {(- 1)}^{r - 1} . x^{r}}{r!} \end{array}$