Chapter 10 Counting Methods and Probability - 10.2 Use Combinations and Binomial Theorem - 10.2 Exercises - Skill Practice - Page 695: 47

See below.

We know that $_nC_r=\frac{n!}{r!(n-r)!}$. Hence $_{n+1}C_r=\frac{(n+1)!}{r!(n+1-r)!}$ $_{n}C_r+_{n}C_{r-1}=\frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)!}=\frac{n!(n-r+1)}{r!(n-r+1)!}+\frac{n!r}{r!(n-r+1)!}=\frac{n!(n-r+1)+n!r}{r!(n-r+1)!}==\frac{(n+1)!}{r!(n+1-r)!}$ Thus they are equal.