13C0−13C1+13C2−........+13C12

Given series 13C0−13C1+13C2−........+13C12   We know that expansion of (1+x)n is (1+x)n=nC0+x⋅nC1+x2⋅nC2+....+xr⋅nCr+...+xn⋅nCn           Here nC0=C0,nC1=C1......,nCr=Cr...nCn=Cn                               Then we have the series of C0−C1+C2−C3+....+(−1)nCn=0 (or) ∑r=0n(−1)r⋅Cr=0 ∴13C0−13C1+13C2−........+13C12   adding and subtracting with 13C13 Then we get ⇒(13C0−13C1+13C2−........+13C12−13C13)  +13C13 ⇒0+1(nCn=1) (0 be the from above series )             =1