Binomial TheoremMedium
Question
Given positive integers r > 1, n > 2and the coefficient of (3r) th and (r + 2) th term in the binomial expansion of (1 + x)2n are equal. Then,
Options
A.n = 2r
B.n = 2r +1
C.n = 3r
D.None of these
Solution
In the expansion (1 + x)2n, t3r = 2nC3r-1 (x)3r-1
and tr+2 = nCr+1 (x)r-1
Since, binomial coefficients of t3r and tr+2 are equal.
⇒ 2nC3r-1 = 2nCr+1
⇒ 3r - 1 = r + 1 or 2n = (3r - 1) + (r + 1)
⇒ 2r = 2 or 2n = 4r
⇒ r =1 or n = 2r
But r > 1,
∴ We take n = 2r
and tr+2 = nCr+1 (x)r-1
Since, binomial coefficients of t3r and tr+2 are equal.
⇒ 2nC3r-1 = 2nCr+1
⇒ 3r - 1 = r + 1 or 2n = (3r - 1) + (r + 1)
⇒ 2r = 2 or 2n = 4r
⇒ r =1 or n = 2r
But r > 1,
∴ We take n = 2r
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