If S=∑i=1n  nCi−1 nCi+nCi−13=3613 then n is equal to

nCr−1+nCr=n+1Cn∴ nCr−1 nCr−1+nCr=n!(r−1)!(n−r+1)!⋅r!(n+1−r)!(n+1)!=rn+1⇒S=∑r=1n rn+13=1(n+1)3⋅n2(n+1)24=n24(n+1)=3613⇒13n2=144(n+1)⇒n=12,−12/13Thus, n=12