Binomial TheoremHard
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
Create a free account to view solution
View Solution FreeMore Binomial Theorem Questions
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 ...If recursion polynomials Pk(x) are defined as P1(x) = (x - 2)2, P2 (x) = ((x - 2)2 - 2)2 P3 (x) = ((x - 2)2 - 2)2 - 2)2 ...If C0, C1, C2.......Cn are binomial coefficients in the expansion of (1 + x)n, n ∈ N, then is equal to-...If the coefficient of $x$ in the expansion of $\left( ax^{2} + bx + c \right)(1 - 2x)^{26}$ is -56 and the coefficients ...The value of +.......... + (-1)n is :...