simplify n+1Ci = n-1Ci

Question

the question is on combinations

Answer 1

ⁿC_r=n!/(r!(n-r)!) so:

ⁿ⁺¹C_i=(n+1)!/(i!(n+1-i)! and ^n-1C_i=(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⁺¹C_i=[n(n+1)(n-1)!]/[i!(n-1-i)!(n-i)(n+1-i)]=n(n+1)^n-1C_i/[(n-i)(n+1-i)].

ⁿ⁺¹C_i=^n-1C_i⇒n(n+1)^n-1C_i/[(n-i)(n+1-i)]=^n-1C_i,

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. ⁿC_r 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.