If the coefficients of the r th, (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

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m2−m(4r+1)+4r2−2=0

m2−m(4r−1)+4r2+2=0

m2−m(4r−1)+4r2−2=0

m2−m(4r+1)+4r2+2=0

answer is A.

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Detailed Solution

Coefficient of the r th term is mCr−1 According to the given condition mCr−1+mCr+1=2 mCr⇒ mCr−1 mCr+ mCr+1 mCr=2⇒rm+1−r+m−rr+1=2⇒ 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)+21−r2⇒ m2−m(4r+1)+4r2−2=0.

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