If Σr=0⁵ (¹¹ C₂ᵣ₊₁)/(2 r+2)=m/n, gcd(m, n)=1, then m-n is equal to __________.

Solution

Maths sir · 4 steps

Absorb the denominator into the coefficient: turn ¹¹Cₖ/(k+1) into ¹²Cₖ₊₁/12, then sum.

1. Absorb the denominator. Here 2r+2 = (2r+1)+1, so each term has the form ¹¹Cₖ/(k+1) with k = 2r+1. Multiply the top and bottom by 12 and the factorials combine into a single coefficient of 12.

   ¹¹Cₖ/(k+1) = 11!/((k+1)! (11-k)!) = 1/12·12!/((k+1)! (11-k)!) = ¹²Cₖ₊₁/12

2. Rewrite the sum. With k+1 = 2r+2 running over 2, 4, …, 12, the sum becomes 1/12(¹²C₂ + ¹²C₄ + … + ¹²C₁₂). That is every even-index coefficient of 12 except ¹²C₀.
3. Even coefficients. All even-index coefficients of 12 add to 2¹¹ = 2048. Removing ¹²C₀ = 1 leaves 2047, so the sum is 2047/12.
4. So. 2047 = 23 × 89 shares no factor with 12 = 2² × 3, so m = 2047 and n = 12. Then m - n = 2035.

The common slip: 509. Forgetting to remove ¹²C₀ gives 2048/12 = 512/3, so m - n = 509, not 2035.

The idea to keep. Whenever ⁿCₖ is divided by k+1, rewrite it as ⁿ⁺¹Cₖ₊₁/(n+1). Then check which terms of the new sum are missing, like ⁿ⁺¹C₀.

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