NDA Mathematics Question (2026) — Solution

Step-by-step solution

Given ( 2 +1)^ 10 = u + f and ( 2 -1)^ 10 = v. Adding the two equations: u + f + v = ( 2 +1)^ 10 + ( 2 -1)^ 10 Using the binomial expansion (x+y)^n + (x-y)^n = 2 [^ n C_ 0 x^n + ^ n C_ 2 x^ n-2 y^2 + ]: u + f + v = 2 [^ 10 C_ 0 ( 2 )^ 10 + ^ 10 C_ 2 ( 2 )^8 + ^ 10 C_ 4 ( 2 )^6 + ^ 10 C_ 6 ( 2 )^4 + ^ 10 C_ 8 ( 2 )^2 + ^ 10 C_ 10 ] u + f + v = 2 [1(32) + 45(16) + 210(8) + 210(4) + 45(2) + 1(1)] u + f + v = 2 [32 + 720 + 1680 + 840 + 90 + 1] u + f + v = 2 [3363] = 6726 Since 0 We are given 0 Since u and 6726 are integers, f + v must be an integer. The only integer strictly between 0 and 2 is 1. Therefore, f + v = 1. Substituting this back into the equation: u + 1 = 6726 u = 6725 Answer: 6725