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
$$\binom{m+n}{r}=\binom{m}{0}\binom{n}{r}+...+\binom{m}{r}\binom{n}{0}$$
 A: HINT:
The coefficient of $x^r,y^r$ in $$\sum_{s=0}^k(x+y)^{r+s}$$ which is a Geometric Series,
then set $x=1$ and $y=1$ one by one
A: $\bf{My\; Solution::}$ Given $\displaystyle \binom{r}{r}+\binom{r+1}{r}+\binom{r+2}{r}+\cdot \cdot \cdot \cdot \cdot \cdot\cdot +\binom{n}{r} = \binom{n+1}{r+1}\;,$
Where $n\geq r \geq 1.$
Coefficeint of $x^r$ in
$\displaystyle \left\{(1+x)^{r}+(1+x)^{r+1}+(1+x)^{r+1}\cdot \cdot \cdot \cdot \cdot \cdot \cdot+(1+x)^{n}\right\} = \frac{(1+x)^{n+1}-(1+x)^r}{(1+x)-1}=\frac{(1+x)^{n+1}-(1+x)^r}{x}$
(Using the formula $\displaystyle \left (r+r^2+..............+r^{n} = \frac{r^{n+1}-1}{r-1} \right)$
So Coeff. of $x^{r}$ in $\displaystyle \frac{(1+x)^{n+1}-(1+x)^r}{x}$
So Coeff. of $x^{r+1}$ in $\displaystyle (1+x)^{n+1}-(1+x)^r = \binom{n+1}{r+1}$
A: $\bf{My\; Solution::}$ 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 $\displaystyle \left\{(1+x)^{r+k+1}-(1+x)^{r}\right\} = \binom{r+k+1}{r+1} = \binom{r+k+1}{k}$ 
