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If the coefficients of rth, (r+1)th and (r+2)th terms in the binomial expansion of (1+y)m are in A.P., then m and r satisfy the equation
- a)m2 - m(4r + 1) + 4r2 - 2 = 0
- b)m2 - m(4r - 1) + 4r2 + 2 = 0
- c)m2 - m(4r - 1) + 4r2 - 2 = 0
- d)m2 - m(4r + 1) + 4r2 + 2 = 0
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
If the coefficients of rth, (r+1)th and (r+2)th terms in the binomial ...
Coefficient of the rth term is mCr - 1
According to the given condition mCr - 1 + mCr + 1 = 2(mCr)
⇒ r2 + r + (m - r) (m + 1 - r) = 2(m + 1 - r) (r + 1)
⇒ r2 + r + m2 + (1 - 2r) m - r(1 - r) = 2m (r + 1) + 2(1 - r2)
⇒ m2 - m(4r + 1) + 4r2 - 2 = 0
According to the given condition mCr - 1 + mCr + 1 = 2(mCr)
⇒ r2 + r + (m - r) (m + 1 - r) = 2(m + 1 - r) (r + 1)
⇒ r2 + r + m2 + (1 - 2r) m - r(1 - r) = 2m (r + 1) + 2(1 - r2)
⇒ m2 - m(4r + 1) + 4r2 - 2 = 0
Most Upvoted Answer
If the coefficients of rth, (r+1)th and (r+2)th terms in the binomial ...
To solve this problem, let's first understand the binomial expansion of (1-y)^m. The general term in the expansion is given by:
T(r) = C(m,r) * (1)^r * (-y)^(m-r)
Where C(m,r) represents the binomial coefficient, given by C(m,r) = m! / (r!(m-r)!)
Now, let's consider the coefficients of the rth, (r+1)th, and (r+2)th terms in the expansion. These coefficients are given by C(m,r), C(m,r+1), and C(m,r+2), respectively.
According to the problem, these coefficients are in an arithmetic progression (A.P.). This means that the common difference between consecutive terms is the same. Therefore, we can write:
C(m,r+1) - C(m,r) = C(m,r+2) - C(m,r+1)
Let's expand these expressions using the formula for the binomial coefficient:
(m! / ((r+1)!(m-r-1)!)) - (m! / (r!(m-r)!)) = (m! / ((r+2)!(m-r-2)!)) - (m! / ((r+1)!(m-r-1)!))
Simplifying this equation, we get:
m! / ((r+1)!(m-r-1)!) - m! / (r!(m-r)!) = m! / ((r+2)!(m-r-2)!) - m! / ((r+1)!(m-r-1)!)
Multiplying through by (r!(m-r)!(r+1)!(m-r-1)!((r+2)!(m-r-2)!), we get:
(r!(m-r)!) - (r!(m-r)(r+2)(m-r-1)) = ((r+1)!(m-r-1)!) - ((r+1)!(m-r-1)(m-r-1)!((r+2)!(m-r-2)!)
Simplifying further, we obtain:
r!(m-r)! - r!(m-r)(r+2)(m-r-1) = (r+1)!(m-r-1)! - (r+1)!(m-r-1)(r+2)(m-r-2)!
Expanding the factorials, we have:
r!(m-r)(m-r-1) - r!(m-r)(r+2)(m-r-1) = (r+1)!(m-r-1)(m-r-2) - (r+1)!(m-r-1)(r+2)
Simplifying this equation, we get:
r!(m-r)(m-r-1 - (r+2)) = (r+1)!(m-r-1)(m-r-2 - (r+2))
r!(m-r)(m-r-1 - r - 2) = (r+1)!(m-r-1)(m-r-2 - r - 2)
r!(m-r)(m-3r-3) = (r+1)!(m-r-1)(m-3r-4)
Dividing both sides by r!(m-r-1), we obtain:
(m-r)(m-3
T(r) = C(m,r) * (1)^r * (-y)^(m-r)
Where C(m,r) represents the binomial coefficient, given by C(m,r) = m! / (r!(m-r)!)
Now, let's consider the coefficients of the rth, (r+1)th, and (r+2)th terms in the expansion. These coefficients are given by C(m,r), C(m,r+1), and C(m,r+2), respectively.
According to the problem, these coefficients are in an arithmetic progression (A.P.). This means that the common difference between consecutive terms is the same. Therefore, we can write:
C(m,r+1) - C(m,r) = C(m,r+2) - C(m,r+1)
Let's expand these expressions using the formula for the binomial coefficient:
(m! / ((r+1)!(m-r-1)!)) - (m! / (r!(m-r)!)) = (m! / ((r+2)!(m-r-2)!)) - (m! / ((r+1)!(m-r-1)!))
Simplifying this equation, we get:
m! / ((r+1)!(m-r-1)!) - m! / (r!(m-r)!) = m! / ((r+2)!(m-r-2)!) - m! / ((r+1)!(m-r-1)!)
Multiplying through by (r!(m-r)!(r+1)!(m-r-1)!((r+2)!(m-r-2)!), we get:
(r!(m-r)!) - (r!(m-r)(r+2)(m-r-1)) = ((r+1)!(m-r-1)!) - ((r+1)!(m-r-1)(m-r-1)!((r+2)!(m-r-2)!)
Simplifying further, we obtain:
r!(m-r)! - r!(m-r)(r+2)(m-r-1) = (r+1)!(m-r-1)! - (r+1)!(m-r-1)(r+2)(m-r-2)!
Expanding the factorials, we have:
r!(m-r)(m-r-1) - r!(m-r)(r+2)(m-r-1) = (r+1)!(m-r-1)(m-r-2) - (r+1)!(m-r-1)(r+2)
Simplifying this equation, we get:
r!(m-r)(m-r-1 - (r+2)) = (r+1)!(m-r-1)(m-r-2 - (r+2))
r!(m-r)(m-r-1 - r - 2) = (r+1)!(m-r-1)(m-r-2 - r - 2)
r!(m-r)(m-3r-3) = (r+1)!(m-r-1)(m-3r-4)
Dividing both sides by r!(m-r-1), we obtain:
(m-r)(m-3
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If the coefficients of rth, (r+1)th and (r+2)th terms in the binomial expansion of (1+y)m are in A.P., then m and r satisfy the equationa)m2 - m(4r + 1) + 4r2 - 2 = 0b)m2 - m(4r - 1) + 4r2 + 2 = 0c)m2 - m(4r - 1) + 4r2 - 2 = 0d)m2 - m(4r + 1) + 4r2 + 2 = 0Correct answer is option 'A'. Can you explain this answer?
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If the coefficients of rth, (r+1)th and (r+2)th terms in the binomial expansion of (1+y)m are in A.P., then m and r satisfy the equationa)m2 - m(4r + 1) + 4r2 - 2 = 0b)m2 - m(4r - 1) + 4r2 + 2 = 0c)m2 - m(4r - 1) + 4r2 - 2 = 0d)m2 - m(4r + 1) + 4r2 + 2 = 0Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If the coefficients of rth, (r+1)th and (r+2)th terms in the binomial expansion of (1+y)m are in A.P., then m and r satisfy the equationa)m2 - m(4r + 1) + 4r2 - 2 = 0b)m2 - m(4r - 1) + 4r2 + 2 = 0c)m2 - m(4r - 1) + 4r2 - 2 = 0d)m2 - m(4r + 1) + 4r2 + 2 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the coefficients of rth, (r+1)th and (r+2)th terms in the binomial expansion of (1+y)m are in A.P., then m and r satisfy the equationa)m2 - m(4r + 1) + 4r2 - 2 = 0b)m2 - m(4r - 1) + 4r2 + 2 = 0c)m2 - m(4r - 1) + 4r2 - 2 = 0d)m2 - m(4r + 1) + 4r2 + 2 = 0Correct answer is option 'A'. Can you explain this answer?.
If the coefficients of rth, (r+1)th and (r+2)th terms in the binomial expansion of (1+y)m are in A.P., then m and r satisfy the equationa)m2 - m(4r + 1) + 4r2 - 2 = 0b)m2 - m(4r - 1) + 4r2 + 2 = 0c)m2 - m(4r - 1) + 4r2 - 2 = 0d)m2 - m(4r + 1) + 4r2 + 2 = 0Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If the coefficients of rth, (r+1)th and (r+2)th terms in the binomial expansion of (1+y)m are in A.P., then m and r satisfy the equationa)m2 - m(4r + 1) + 4r2 - 2 = 0b)m2 - m(4r - 1) + 4r2 + 2 = 0c)m2 - m(4r - 1) + 4r2 - 2 = 0d)m2 - m(4r + 1) + 4r2 + 2 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the coefficients of rth, (r+1)th and (r+2)th terms in the binomial expansion of (1+y)m are in A.P., then m and r satisfy the equationa)m2 - m(4r + 1) + 4r2 - 2 = 0b)m2 - m(4r - 1) + 4r2 + 2 = 0c)m2 - m(4r - 1) + 4r2 - 2 = 0d)m2 - m(4r + 1) + 4r2 + 2 = 0Correct answer is option 'A'. Can you explain this answer?.
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