Binomiaaal

n = 2m+1
The given sum
1^2m+1C₀ - 1^2m+1C₁ + 1^2m+1C₂ - 1^2m+1C₃ + . . . + 1^2m+1C_2m - 1^2m+1C_2m+1

We note that for each k = 0,1,2, ...., 2m+1, the term 1^2m+1C_k has a corresponding term 1^2m+1C_(2m+1)-k with opposite sign. For example, for 1^2m+1C₀ we have -1^2m+1C_2m+1; for -1^2m+1C₁ we have 1^2m+1C_2m and so on.

Since total number of terms is even, we see that pairwise terms cancel out. As a result the sum is zero.