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

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Binomial Theorem

Let the coefficients of

x^{- 1}

and

x^{- 3}

in the expansion of

(2 x^{\frac{1}{5}} - \frac{1}{x ^{\frac{1}{5}}})^{15}, x > 0

, be

m

and

n

respectively. If r is a positive integer such that

m n^{2} =^{15} C_{r} . 2^{r}

, then the value of

r

is equal to ______.

Answer: 5

The general term in the binomial expansion is:

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

For the coefficient of

x^{- 1}

\frac{15 - 2 r}{5} = - 1

gives

15 - 2 r = - 5

, so

r = 10

.
Thus

m = (10 15) 2^{5} = (5 15) 2^{5}

.

For the coefficient of

x^{- 3}

\frac{15 - 2 r}{5} = - 3

gives

15 - 2 r = - 15

, so

r = 15

.
Thus

n = (- 1)^{15} (15 15) 2^{0} = - 1

.

Therefore:

m n^{2} = (5 15) 2^{5} \cdot 1 = (5 15) 2^{5}

Comparing with

(r 15) 2^{r}

, we get

r = 5

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Integer answer