Answer: The proof is given below.
Step-by-step explanation: We are given to show that the following equality is true :
[tex]^{n+1}C_r=^nC_{r-1}+^nC_r.[/tex]
We know that
the number of combinations of n different things taken r at a time is given by
[tex]^nC_r=\dfrac{n!}{r!(n-r)!}.[/tex]
Therefore, we have
[tex]R.H.S.\\\\=^nC_{r-1}+^nC_r\\\\\\=\dfrac{n!}{(r-1)!(n-(r-1))!}+\dfrac{n!}{(r)!(n-r)!}\\\\\\=\dfrac{n!}{(r-1)!(n-r+1)!}+\dfrac{n!}{(r)!(n-r)!}\\\\\\=\dfrac{n!}{(r-1)!(n-r+1)(n-r)!}+\dfrac{n!}{r(r-1)!(n-r)!}\\\\\\=\dfrac{n!}{(r-1)!(n-r)!}\left(\dfrac{1}{n-r+1}+\dfrac{1}{r}\right)\\\\\\=\dfrac{n!}{(r-1)!(n-r)!}\left(\dfrac{r+n-r+1}{(n-r+1)r}\right)\\\\\\=\dfrac{n!}{(r-1)!(n-r)!}\times\dfrac{n+1}{(n-r+1)r}\\\\\\=\dfrac{(n+1)!}{r!(n-r+1)!}\\\\\\=\dfrac{(n+1)!}{r!((n+1)-r)!}\\\\\\=^{n+1}C_r\\\\=L.H.S.[/tex]
Thus, [tex]^{n+1}C_r=^nC_{r-1}+^nC_r.[/tex]
Hence proved.
