JEE Main 2020 · 4 September · evening · Single correct · +4 / −1

If for some positive integer n, the coefficients
of three consecutive terms in the binomial
expansion of (1 + x) n + 5 are in the ratio
5 : 10 : 14, then the largest coefficient in this expansion is :

  1. (a) 330
  2. (b) 792
  3. (c) 252
  4. (d) 462
Answer and solution

Answer (d) 462

Solution

Maths sir · 4 steps

Turn the ratio into two ratios of neighbouring coefficients, find the power, then pick the middle term.

  1. Set up. Let N = n + 5 and let the three coefficients be C(N, r-1), C(N, r), C(N, r+1). Neighbouring coefficients follow a simple rule: C(N, r)/C(N, r-1) = (N-r+1)/r.

    (N-r+1)/r = 10/5 = 2, (N-r)/(r+1) = 14/10 = 7/5

  2. Solve the pair. The first gives N = 3r - 1. The second gives 5N - 5r = 7r + 7, so 5N = 12r + 7. Substituting, 15r - 5 = 12r + 7, so r = 4 and N = 11, which means n = 6.
  3. Check. C(11, 3) = 165, C(11, 4) = 330, C(11, 5) = 462. Dividing each by 33 gives 5 : 10 : 14, as required.
  4. So. For (1+x)¹¹ the power is odd, so the two middle coefficients C(11, 5) = C(11, 6) are the largest. The largest coefficient is 462, option (d).

The idea to keep. For consecutive binomial coefficients, use C(N, r)/C(N, r-1) = (N-r+1)/r. Two such ratios give two equations in N and r. The largest coefficient always sits at the middle term.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Binomial Theorem

Asked about 200 times in JEE Main since 2020. This question is from 2020.