Question

Let the coefficients of x^–1 and x^–3 in the expansion of
\((2x^{\frac{1}{5}} - \frac{1}{x^{\frac{1}{5}}} )^{15} , x > 0\)be m and n respectively. If r is a positive integer such that
\(mn² = ^{15}C_r.2^r\)
then the value of r is equal to ______.

Answer 1

Correct Answer: 5

Consider the expansion of \( (2x^{\frac{1}{5}} - \frac{1}{x^{\frac{1}{5}}})^{15} \). The general term is given by:

T_k = \binom{15}{k} (2x^{\frac{1}{5}})^{15-k} \left(-\frac{1}{x^{\frac{1}{5}}}\right)^{k}

Simplifying \( T_k \), we have:

\(T_k = \binom{15}{k} \cdot 2^{15-k} \cdot (-1)^k \cdot x^{\frac{1}{5}(15-k)} \cdot x^{-\frac{k}{5}}\)

\( = \binom{15}{k} \cdot 2^{15-k} \cdot (-1)^k \cdot x^{\frac{1}{5}(15-k-k)}\)

\( = \binom{15}{k} \cdot 2^{15-k} \cdot (-1)^k \cdot x^{\frac{15-2k}{5}}\)

For the coefficient of \( x^{-1} \):

\( \frac{15-2k}{5} = -1 \)

\( 15-2k = -5 \)

\( 2k = 20 \)

\( k = 10 \)

Coefficient of \( x^{-1} \) is:

\(\binom{15}{10} \cdot 2^{5} \cdot (-1)^{10} = \binom{15}{10} \cdot 32\)

For the coefficient of \( x^{-3} \):

\( \frac{15-2k}{5} = -3 \)

\( 15-2k = -15 \)

\( 2k = 30 \)

\( k = 15 \)

Coefficient of \( x^{-3} \) is:

\(\binom{15}{15} \cdot 2^{0} \cdot (-1)^{15} = -1\)

Given \( mn^2 = \binom{15}{r} \cdot 2^r \), find \( r \):

\( m = \binom{15}{10} \cdot 32 \) and \( n = -1 \)

\( (-1)^2 = 1 \)

\( mn^2 = \binom{15}{10} \cdot 32 \)

\( \binom{15}{r} \cdot 2^r = \binom{15}{10} \cdot 32 \)

\( 32 = 2^5 \) implies \( r = 5 \)

Checking \( r \) in range 5, 5: correct. Thus, \( r = 5 \).