Question

4) In the expansion of (x-2x)", there will be a constant term with Xo. Find the product of the smallest natural numberin) and the value of this constant.

Answer 1

Answer:-

Given that:-

In the expansion of $(x-\frac{1}{2\sqrt{x}})^{n}=(x-\frac{1}{2x^{1/3}})^{n}$

term is given by,
$(-1)^{r}n_{C_{r}}x^{n-r}(\frac{1}{2x^{1/3}})^{r}$

$=(-1)^{r}n_{C_{r}}x^{n-r}\frac{1}{2^{r}x^{r/3}}$

$=(-1)^{r}\, \, n_{C_{r}}\, \, x^{n-r-\frac{r}{3}}\, \, \, 2^{-r}$

$=(-1)^{r}\, \, n_{C_{r}}\, \,x^{\frac{3n-4r}{3}}\, \, 2^{-r}$

$x^{\frac{3n-4r}{3}}=x^{0}$

$\Rightarrow \frac{3n-4r}{3}=0$

$\Rightarrow 3n^{3}=4r$

$\Rightarrow r=\frac{3n}{4}$

Since r is a positive integer the minimum value of n (natural number)

for which r will be a positive integer.

is, n=4  $\Rightarrow \begin{bmatrix} for\, \, n=1,\frac{3}{4}\notin |N, \\ for\, \, n=2,\frac{6}{4}\notin |N,for\, \, n=3,\frac{9}{4}\notin |N, \end{bmatrix}$

So, n=4 and $r=\frac{3\times 9}{9}=3$

The constant term is,

$(-1)^{3}4_{C_{3}}2^{-3}$

$=(-1)4.\frac{1}{8}=-\frac{1}{2}$

The product of n and $(-\frac{1}{2})$ is,

$4\times (-\frac{1}{2})$

$=-2$

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