Question

find the coefficient of x in the expansion of (1-2x3+3x5)(1+1/x)8 ​​

Answer 1

Please see the attachment

Answer 2

The coefficient of $x$ is $154$.

Step-by-step explanation:

Given,

Find the coefficient of $x$ in the expansion of $(1-2x^{3}+3x^{5})(1+\frac{1}{x})^{8}$

∴  $(1-2x^{3}+3x^{5})(1+\frac{1}{x})^{8}$

From above expression,

We will find coefficient of $x^{1},x^{-2}  and x^{-4}$.

So we have,

$T_{r+1} =_{r}^{8}\textrm{\mathbb{C}}(1)^{8-r}(\frac{1}{x})^{r}$

⇒$T_{r+1} =_{r}^{8}\textrm{\mathbb{C}}(1)^{8-r}x^{-r}$

∴ $x^{-r}=x^{1}$

⇒$r=-1$

This value is not possible.

And, $x^{-r}=x^{-2}$

⇒$r=2$

Also,

$x^{-r}=x^{-4}$

⇒$r=4$

So, For $r=2$ ,

$T_{2+1} =-2\times _{2}^{8}\textrm{\mathbb{C}}(1)^{8-2}$

⇒$T_{3} =-2\times \frac{8!}{2!\times (8-2)!}$

⇒$T_{3} =-2\times \frac{8\times 7}{2}$

⇒$T_{3} =-56$

And, $r=4$,

$T_{4+1} =3\times _{4}^{8}\textrm{\mathbb{C}}(1)^{8-4}$

⇒$T_{5} =3\times \frac{8!}{4!\times (8-4)!}$

⇒$T_{5} =3\times \frac{8\times 7\times 6\times 5}{4\times 3\times 2}$

⇒$T_{5} =210$

∴ $T_{3}+T_{5}$

$=-56+210$

$=154$

So, The coefficient of $x$ is $154$.