JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The coefficient of x^3 in the given expression is the sum of the coefficients of x^3 in each term. The coefficient of x^3 in _ r=3 ^ 99 (1+x)^r + (1+kx)^ 100 is: _ r=3 ^ 99 ^ r C_3 + k^3 ^ 100 C_3 Using the identity _ r=k ^ n ^ r C_k = ^ n+1 C_ k+1 , we get: _ r=3 ^ 99 ^ r C_3 = ^ 100 C_4 Thus, the total coefficient of x^3 is ^ 100 C_4 + k^3 ^ 100 C_3. Given that this coefficient is equal to (43n + 101 4 ) ^ 100 C_3, we have: ^ 100 C_4 + k^3 ^ 100 C_3 = (43n + 101 4 ) ^ 100 C_3 Dividing both sides by ^ 100 C_3: ^ 100 C_4 ^ 100 C_3 + k^3 = 43n + 101 4 Using ^ n C_r ^ n C_ r-1 = n-r+1 r , we get ^ 100 C_4 ^ 100 C_3 = 100-4+1 4 = 97 4 . Substituting this value: 97 4 + k^3 = 43n + 101 4 k^3 = 43n + 101 4 - 97 4 k^3 = 43n + 1 k^3 - 1 = 43n Since n N , we must have n 1, which implies k^3 - 1 43 k 4. Checking the values of k 4 to find the smallest integer k such that k^3 - 1 is divisible by 43: For k = 4, k^3 - 1 = 63 (not divisible by 43). For k = 5, k^3 - 1 = 124 (not divisible by 43). For k = 6, k^3 - 1 = 215 = 43 5. Thus, the smallest value of k is p = 6, and the corresponding value of n is 5. Therefore, p + n = 6 + 5 = 11. Answer: 11