Proving Identity of Combination The lecturer had given two questions of proving that are
$$\binom{r}{r}+\binom{r+1}{r}+...+\binom{n}{r}=\binom{n+1}{r+1}\text{for }n\geq{r}\geq{1}   $$
$$\binom{r}{0}+\binom{r+1}{1}+...+\binom{r+k}{k}=\binom{r+k+1}{k}\text{for }r,k\geq{1}$$
I tried to use the induction to prove these two identites but the lecturer said these two proving questions should be related to the identity which is
$$\binom{m+n}{r}=\binom{m}{0}\binom{n}{r}+...+\binom{m}{r}\binom{n}{0}$$

$\bf{My\; Solution::}$ Given 
$\displaystyle \binom{r}{0}+\binom{r+1}{1}+\binom{r+2}{2}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot +\binom{r+k}{k} = \binom{r+k+1}{k}\;,$ where $r,k\geq 1$
Now Using The formula of $\displaystyle \binom{n}{r} = \binom{n}{n-1}$ on $\bf{L.H.S\;}$ 
We can Write $\bf{L.H.S}$ as 
$\displaystyle \binom{r}{r}+\binom{r+1}{r}+\binom{r+2}{r}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot +\binom{r+k}{r}$
Coeff. of $x^r$ in $\displaystyle \left\{(1+x)^r+(1+x)^{r+1}+(1+x)^{r+2}+\cdot \cdot \cdot \cdot +(1+x)^{r+k}\right\} = \frac{(1+x)^{r+k+1}-(1+x)^{r}}{(1+x)-1}$
So Coeff. of $x^{r+1}$ in $\displaystyle \left\{(1+x)^{r+k+1}-(1+x)^{r}\right\} = \binom{r+k+1}{r+1} = \binom{r+k+1}{k}$ 
