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2001

JEE Main 2001 Mathematics question on Binomial Theorem

In the binomial expansion of (a−b)n(a - b)^n, n≥5n \geq 5 the sum of the 5th and 6th terms is zero. Then, a/ba/b equals
  1. A.
    n−56\dfrac{n-5}{6}
  2. B.
    n−45\dfrac{n-4}{5}
  3. C.
    5n−4\dfrac{5}{n-4}
  4. D.
    6n−5\dfrac{6}{n-5}
Show answer and solution

Correct option: B

Solution

The 5th term is C(n,4)·a^(n-4)·(-b)4 = C(n,4)·a^(n-4)·b4 and the 6th term is C(n,5)·a^(n-5)·(-b)5 = -C(n,5)·a^(n-5)·b5.

Setting their sum to zero: C(n,4)·a^(n-4)·b4 = C(n,5)·a^(n-5)·b5.

Dividing both sides by a^(n-5)·b4: C(n,4)·a = C(n,5)·b, so a/b = C(n,5)/C(n,4) = (n-4)/5.

[2001]

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