If (¹¹ C₁)/2+(¹¹ C₂)/3+…+(¹¹ C₉)/10=n/m with gcd(n, m)=1 , then n+m is equal to _______.

Solution

Maths sir · 4 steps

Turn each term into a ¹²C coefficient, and the sum becomes a full row with its ends cut off.

1. Absorb the denominator. Use 1/(r+1) ¹¹Cᵣ = 1/12 ¹²Cᵣ₊₁. You can check it by writing both sides with factorials: each equals 11!/((r+1)! (11-r)!).

   Σᵣ₌₁⁹¹¹Cᵣ/(r+1) = 1/12(¹²C₂ + ¹²C₃ + … + ¹²C₁₀)

2. Use the full row. All the coefficients of the row add to 2¹² = 4096. Remove the missing ends ¹²C₀ = 1, ¹²C₁ = 12, ¹²C₁₁ = 12 and ¹²C₁₂ = 1. That leaves 4096 - 26 = 4070.
3. Reduce the fraction. The sum is 4070/12 = 2035/6. Here 2035 is odd and its digit sum is 10, so neither 2 nor 3 divides it. That gives gcd(2035, 6) = 1.
4. So. n = 2035 and m = 6, so n + m = 2041.

The common slip: 4082. Stopping at 4070/12 without cancelling the common factor 2 gives 4070 + 12 = 4082, not 2041.

The idea to keep. When ⁿCᵣ is divided by r+1, rewrite it as 1/(n+1) ⁿ⁺¹Cᵣ₊₁. Then sum the whole row as 2ⁿ⁺¹ and subtract the terms that are not there.

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