nCr+2nCr−1+nCr−2 is equal to
n+1Cr
nCr−1
n−1Cr−1
n+2Cr
We have
nCr+2nCr−1+nCr−2= nCr+nCr+1+ nCr−j+nCr−2=n+1Cr+n+1Cr−1=n+2Cγ