nCr=n!/(r!(n-r)!) so:
n+1Ci=(n+1)!/(i!(n+1-i)! and n-1Ci=(n-1)!/(i!(n-1-i)!.
(n+1)!=n(n+1)(n-1)! and (n+1-i)!=(n-1-i)!(n-i)(n+1-i).
n+1Ci=[n(n+1)(n-1)!]/[i!(n-1-i)!(n-i)(n+1-i)]=n(n+1)n-1Ci/[(n-i)(n+1-i)].
n+1Ci=n-1Ci⇒n(n+1)n-1Ci/[(n-i)(n+1-i)]=n-1Ci,
n(n+1)/[(n-i)(n+1-i)]=1,
n2+n=(n-i)(n+1-i),
n2+n=n2+n-in-in-i+i2,
0=-2in-i+i2,
0=-2n-1+i (for i≠0),
i=2n+1. Note that n≥2 because n-1≥1. nCr requires r≤n, but 2n+1>n+1 (n>0) and 2n+1>n-1 (n>-2), so i≠2n+1.
The alternative is i=0 and is the solution.