For a positive integer n, (1 + (1)/(x)) ^ n is expanded in increasing powers of x . If three consecutive coefficients in this expansion are in the ratio, 2: 5: 12, then n is equal to

Correct answer: 0 Cr - 1 n : Cr n : Cr + 1 n = 2: 5: 12 ⇒ (Cr n)/(Cr - 1 n) = (5)/(2) and (Cr + 1 n)/(Cr n) = (12)/(5) ⇒ (n - r + 1)/(r) = (5)/(2) and (n - r)/(r + 1) = (12)/(5) ⇒ 2n - 7r + 2 = 0 and 5n - 17r - 12 = 0 On solving n = 118 and r = 34

JEE Main 20202 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 2 September
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2020 · 2 September · Evening Shift · Mathematics · Binomial Theorem

For a positive integer n, (1 + (1)/(x)) ^ n is expanded in...

For a positive integer

n, {\left(1 + \dfrac{1}{x}\right)}^{n} is expanded in increasing powers of

. If three consecutive coefficients in this expansion are in the ratio,

2: 5: 12, then

n

is equal to

Correct answer

Step-by-step solution

$Cr - 1 n : Cr n : Cr + 1 n = 2: 5: 12$

$⇒ \dfrac{Cr n}{Cr - 1 n} = \dfrac{5}{2}$ and $\dfrac{Cr + 1 n}{Cr n} = \dfrac{12}{5}$

$⇒ \dfrac{n - r + 1}{r} = \dfrac{5}{2}$ and $\dfrac{n - r}{r + 1} = \dfrac{12}{5}$

$⇒ 2n - 7r + 2 = 0$ and $5n - 17r - 12 = 0$

On solving $n = 118$ and $r = 34$

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