If ( ⁴⁰C₀ ) + ( ⁴¹C₁ ) + ( ⁴²C₂ ) + ..... + ( ⁶⁰C₂₀ ) = m/n⁶⁰C₂₀ m and n are coprime, then m + n is equal to ___________.

Solution

Maths sir · 4 steps

Swap the first term for ⁴¹C₀ and Pascal's rule folds the whole sum into a single coefficient.

1. Rewrite first term. ⁴⁰C₀ = 1 = ⁴¹C₀, so the sum becomes ⁴¹C₀ + ⁴¹C₁ + ⁴²C₂ + … + ⁶⁰C₂₀.
2. Fold step by step. Pascal's rule ⁿCᵣ₋₁ + ⁿCᵣ = ⁿ⁺¹Cᵣ gives ⁴¹C₀ + ⁴¹C₁ = ⁴²C₁. Then ⁴²C₁ + ⁴²C₂ = ⁴³C₂, and so on. Each step raises both the top and the bottom number by 1.

   Σₖ₌₀²⁰ ⁴⁰⁺ᵏCₖ = ⁶⁰C₁₉ + ⁶⁰C₂₀ = ⁶¹C₂₀

3. Compare with 60C20. ⁶¹C₂₀ = 61!/(20! 41!) = 61/41 × 60!/(20! 40!) = 61/41 ⁶⁰C₂₀.
4. So. m/n = 61/41. 41 is prime and does not divide 61, so the fraction is already in lowest terms. That gives m + n = 61 + 41 = 102.

The common slip: 82. Collapsing the sum to ⁶¹C₂₁ instead of ⁶¹C₂₀ gives 61/21 and so 82. The bottom index stays at 20, the last one in the sum.

The idea to keep. When the top and bottom of ⁿCᵣ rise together, first make the opening term match the next one, then let Pascal's rule collapse the sum: Σₖ₌₀ʳ ⁿ⁺ᵏCₖ = ⁿ⁺ʳ⁺¹Cᵣ.

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