Question

Consider the expansion ${\left(x^2+\frac{1}{x}\right)}^{15}$. What is the ratio of coefficient of $x^{15}$ to the term independent of $x$ in the given expansion?

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{3}{4}$

Answer 1

The correct option is A $1$

The genral term of it can be written as $T_{r+1}{=}^{15}{C}_r.x^{2.(15-r)}.\frac{1}{x^r}{=}^{15}{C}_r.x^{(30-3r)}$

$T_{r+1}^{\prime }$ where power of $x$ is $15$.

$\Rightarrow$ $30-3r=15$

$\Rightarrow$ $r=5$

Hence, coefficient is ${}^{15}{C}_5$

$T_{r+1}^{\prime \prime }$, where coefficient of $x$ is $0$.

$\Rightarrow$ $30-3r=0$

$\Rightarrow$ $r=10$

Hence, coefficient is ${}^{15}{C}_{10}$

$\Rightarrow$$\frac{T_{r+1}^{\prime }}{T_{r+1}^{\prime \prime }}=\frac{{}^{15}{C}_5}{{}^{15}{C}_{10}}$

As we know ${}^n{C}_x{=}^n{C}_{n-x}$

$\Rightarrow$$\frac{T_{r+1}^{\prime }}{T_{r+1}^{\prime \prime }}=1$