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Given positive integers r > 1, n >2 and that the coefficient of (3r)th and (r + 2)th terms in the binomial expansion of (1+ x)2n are equal . Then (1983 - 1 Mark)
- a)n = 2r
- b)n = 2r + 1
- c)n = 3r
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Given positive integers r > 1, n >2 and that the coefficient of ...
Given that r and n are +ve integers such that r > 1, n > 2
Also in the expansion of (1+ x)2n coeff. of (3r)th term = coeff. of (r + 2)th term
⇒ 2nC3r–1= 2nCr+1
⇒ 3r –1 = r + 1 or 3r – 1+ r + 1 = 2n
⇒ r = 1 or 2r = n
But r > 1
∴ n = 2r
Also in the expansion of (1+ x)2n coeff. of (3r)th term = coeff. of (r + 2)th term
⇒ 2nC3r–1= 2nCr+1
⇒ 3r –1 = r + 1 or 3r – 1+ r + 1 = 2n
⇒ r = 1 or 2r = n
But r > 1
∴ n = 2r
Most Upvoted Answer
Given positive integers r > 1, n >2 and that the coefficient of ...
We know that the smallest positive integer that is divisible by both r and s is their least common multiple, denoted by LCM(r,s). So, if r and s are relatively prime (i.e., they have no common factors other than 1), then their LCM will be equal to their product, LCM(r,s) = r*s.
Therefore, if r and s are relatively prime, the smallest positive integer that is divisible by both r and s is r*s.
Therefore, if r and s are relatively prime, the smallest positive integer that is divisible by both r and s is r*s.
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Given positive integers r > 1, n >2 and that the coefficient of (3r)th and (r + 2)th terms in the binomial expansion of (1+ x)2n are equal . Then (1983 - 1 Mark)a)n = 2rb)n = 2r + 1c)n = 3rd)none of theseCorrect answer is option 'A'. Can you explain this answer?
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Given positive integers r > 1, n >2 and that the coefficient of (3r)th and (r + 2)th terms in the binomial expansion of (1+ x)2n are equal . Then (1983 - 1 Mark)a)n = 2rb)n = 2r + 1c)n = 3rd)none of theseCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Given positive integers r > 1, n >2 and that the coefficient of (3r)th and (r + 2)th terms in the binomial expansion of (1+ x)2n are equal . Then (1983 - 1 Mark)a)n = 2rb)n = 2r + 1c)n = 3rd)none of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Given positive integers r > 1, n >2 and that the coefficient of (3r)th and (r + 2)th terms in the binomial expansion of (1+ x)2n are equal . Then (1983 - 1 Mark)a)n = 2rb)n = 2r + 1c)n = 3rd)none of theseCorrect answer is option 'A'. Can you explain this answer?.
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