If _ k=1 ¹⁵ k ^ k+2 C₃ = a b ¹⁸C₄ , where a and b are coprime natural numbers, then the value of a + b is

Step-by-step solution

We can rewrite the term k ^ k+2 C₃ by adding and subtracting 3: k ^ k+2 C₃ = (k+3 - 3) ^ k+2 C₃ = (k+3) ^ k+2 C₃ - 3 ^ k+2 C₃ Using the absorption identity n ^ n-1 C_ r-1 = r ^ n C_ r , we have (k+3) ^ k+2 C₃ = 4 ^ k+3 C₄ . So, k ^ k+2 C₃ = 4 ^ k+3 C₄ - 3 ^ k+2 C₃ . Now, summing from k=1 to 15 : _ k=1 ¹⁵ k ^ k+2 C₃ = 4 _ k=1 ¹⁵ ^ k+3 C₄ - 3 _ k=1 ¹⁵ ^ k+2 C₃ = 4 ( ⁴C₄ + ⁵C₄ + + ¹⁸C₄ ) - 3 ( ³C₃ + ⁴C₃ + + ¹⁷C₃ ) Using the hockey-stick identity _ i=r ^ n ^ i C_ r = ^ n+1 C_ r+1 : = 4 ¹⁹C₅ - 3 ¹⁸C₄ Next, express ¹⁹C₅