Question

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

विकल्प

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Answer 1

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

Explanation:

Given expansion is `(2x^(1/5) - 1/x^(1/2))^15`

Now, the general term of given expansion is

T_p+1 = `(-1)^p  ""^15C_p  2^(15-p)(x^(1/5))^(15-p).(1/x^(1/5))^p`

= `(-1)^p  ""^15C_p  2^(15 - p).x^((15-2p)/5)`

For coefficient of `x^-1, (15 - 2p)/5` = –1

⇒ p = 10

∴ m = ¹⁵C₁₀ 2⁵

For coefficient of `x^-3, (15 - 2p)/5` = –3

⇒ p = 15

∴ n = –¹⁵C₁₅ 2⁰ = –1

Now, mn² = ¹⁵C₁₀ 2⁵

⇒ mn² = ¹⁵C₅ 2⁵

⇒ ¹⁵C_r 2^r = ¹⁵C₅ 2⁵

⇒  r = 5