If (1+x)n=C0+C1x+C2x2+⋯+Cnxn, then C0−C0+C1+C0+C1+C2−C0+C1+C2+C3+⋯(−1)n−1C0+C1+⋯+Cn−1 is (where n is even integer and Cr=nCr)

For n = 2m, the given expression isC0−C0+C1+C0+C1+C2−C0+C1+C2+C3+⋯(−1)n−1C0+C1+⋯+Cn−1=C0−C0+C1+C0+C1+C2−C0+C1+C2+C3+⋯−C0+C1+⋯+C2m−1 =−C1+C3+C5+⋯+C2m−1 =−C1+C3+C5+⋯+Cn−1=−2n−1