If 1/(n+1) ⁿ Cₙ+1/n ⁿ Cₙ₋₁+…+1/2 ⁿ C₁+ ⁿ C₀=1023/10 then n is equal to :

Solution

Maths sir · 4 steps

Turn each 1/(r+1) ⁿCᵣ into ⁿ⁺¹Cᵣ₊₁ over n+1, and the sum collapses.

1. Read the general term. Read from ⁿC₀ upwards, every term has the form 1/(r+1) ⁿCᵣ, for r = 0, 1, …, n.
2. Use a standard identity. Since 1/(r+1) ⁿCᵣ = n!/((r+1)! (n-r)!) = 1/(n+1) ⁿ⁺¹Cᵣ₊₁, the whole sum becomes 1/(n+1)(ⁿ⁺¹C₁ + ⁿ⁺¹C₂ + … + ⁿ⁺¹Cₙ₊₁). The bracket holds all the coefficients of (1+x)ⁿ⁺¹ except ⁿ⁺¹C₀ = 1, so it equals 2ⁿ⁺¹ - 1.

   Σᵣ₌₀ⁿ ⁿCᵣ/(r+1) = (2ⁿ⁺¹-1)/(n+1)

3. Match the value. We need (2ⁿ⁺¹-1)/(n+1) = 1023/10. Since 1023 = 2¹⁰ - 1, taking n + 1 = 10 matches both the numerator and the denominator exactly.
4. So. n = 9, which is option (a). Check: (2¹⁰-1)/10 = 1023/10.

The idea to keep. Whenever ⁿCᵣ is divided by r+1, rewrite it as 1/(n+1) ⁿ⁺¹Cᵣ₊₁, or integrate (1+x)ⁿ from 0 to 1. Both give (2ⁿ⁺¹-1)/(n+1).

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