JEE Main 2023 · 1 February · morning · Single correct · +4 / −1

The value of 1/(1 ! 50 !)+1/(3 ! 48 !)+1/(5 ! 46 !)+… .+1/(49 ! 2 !)+1/(51 ! 1 !) is :

  1. (a) 2⁵¹/(50 !)
  2. (b) 2⁵¹/(51 !)
  3. (c) 2⁵⁰/(50 !)
  4. (d) 2⁵⁰/(51 !)
Answer and solution

Answer (d) 2⁵⁰/(51 !)

Solution

Maths sir · 4 steps

The two factorials in each term add up to 51, so multiply through by 51! to get binomial coefficients.

  1. Spot the pattern. In each term 1/(r! (51-r)!) the two factorials add to 51, and r runs over the odd numbers 1, 3, 5, …, 51. The last term 1/(51! 1!) fits the same pattern because 1! = 0! = 1.
  2. Multiply by 51!. Each term becomes 51!/(r! (51-r)!) = C(51, r). So 51! times the sum is the sum of the odd-index coefficients of (1+x)⁵¹.

    C(51, 1)+C(51, 3)+…+C(51, 51) = 2⁵¹/2 = 2⁵⁰

  3. Why half. All the coefficients add to 2⁵¹. Putting x = -1 shows the odd-index and even-index coefficients have equal sums, so each sum is half of 2⁵¹, which is 2⁵⁰.
  4. So. The value is 2⁵⁰/51!, which is option (d).

The idea to keep. When the factorials in the denominators always add to the same n, multiply by n! to get C(n, r). The odd-index or even-index coefficients alone add to 2ⁿ⁻¹.

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 2023.