Let the coefficients of x^-1 and x^-3 in the expansion of (2x^1/5-1x^1/5)^15, x0, be m an…

Question

Let the coefficients of $x^{-1}$ and $x^{-3}$ in the expansion of $\left(2x^{1/5}-\dfrac{1}{x^{1/5}}\right)^{15}$, $x>0$, be $m$ and $n$ respectively. If $r$ is a positive integer such that $mn^2={}^{15}C_r\cdot2^r$, then the value of $r$ is equal to\_\_\_

Answer: 5

Accepted Answer

$$T_{r+1}=(-1)^r\binom{15}{r}2^{15-r}x^{\frac{15-2r}{5}}$$

For $x^{-1}$: $\dfrac{15-2r}5=-1\Rightarrow r=10$; so $m=\binom{15}{10}2^5$.

For $x^{-3}$: $\dfrac{15-2r}5=-3\Rightarrow r=15$; so $n=-1$.

$$mn^2=\binom{15}{10}2^5\times1={}^{15}C_5\cdot2^5\quad\Rightarrow\quad r=5$$

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