# Find the ratio of the coefficient of

Question:

Find the ratio of the coefficient of $x^{15}$ to the term independent of $x$ in the expansion of $\left(x^{2}+\frac{2}{x}\right)^{15}$

Solution:

To Find: the ratio of the coefficient of $x^{15}$ to the term independent of $x$

Formula : $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$

Here, $a=x^{2}, \quad b=\frac{2}{x}$ and $n=15$

We have a formula,

$\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$

$=\left(\begin{array}{c}15 \\ r\end{array}\right)\left(x^{2}\right)^{15-r}\left(\frac{2}{x}\right)^{r}$

$=\left(\begin{array}{c}15 \\ r\end{array}\right)(x)^{30-2 r}(2)^{r}(x)^{-r}$

$=\left(\begin{array}{c}15 \\ r\end{array}\right)(x)^{30-2 r-r}(2)^{r}$

$=\left(\begin{array}{c}15 \\ r\end{array}\right)(2)^{r}(x)^{30-3 r}$

To get coefficient of $x^{15}$ we must have,

$(x)^{30-3 r}=x^{15}$

- $30-3 r=15$

- $3 r=15$

$\cdot r=5$

Therefore, coefficient of $x^{15}=\left(\begin{array}{c}15 \\ 5\end{array}\right)(2)^{5}$

Now, to get coefficient of term independent of $x$ that is coefficient of $x^{0}$ we must have,

$(x)^{30-3 r}=x^{0}$

$-30-3 r=0$

$\cdot 3 r=30$

$\cdot r=10$

Therefore, coefficient of $x^{0}=\left(\begin{array}{l}15 \\ 10\end{array}\right)(2)^{10}$

But $\left.\left(\begin{array}{l}15 \\ 10\end{array}\right)=\left(\begin{array}{c}15 \\ 5\end{array}\right) \ldots \ldots \ldots\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}\mathrm{n} \\ \mathrm{n}-\mathrm{r}\end{array}\right)\right]$

Therefore, the coefficient of x0 $=\left(\begin{array}{c}15 \\ 5\end{array}\right)(2)^{10}$

Therefore,

$\frac{\text { coefficient of } x^{15}}{\text { coefficient of } x^{0}}=\frac{\left(\begin{array}{c}15 \\ 5\end{array}\right)(2)^{5}}{\left(\begin{array}{c}15 \\ 5\end{array}\right)(2)^{10}}$

$=\frac{1}{(2)^{5}}$

$=\frac{1}{32}$

Hence, coefficient of $x^{15}:$ coefficient of $x^{0}=1: 32$

$\underline{\text { Conclusion }}$ : The ratio of coefficient of $x^{15}$ to coefficient of $x^{0}=1: 32$