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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 $$\binom{m+n}{r}=\binom{m}{0}\binom{n}{r}+...+\binom{m}{r}\binom{n}{0}$$

Wang Kah Lun
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    Find $$\binom{n+1}{r+1}-\binom nr$$ and $$\binom{r+k+1}k-\binom{r+k}k$$ – lab bhattacharjee Nov 02 '14 at 05:54
  • I said already I used these two and by using induction to prove the questions but the lecturer said to me the questions should be related to the Vandermonde identity. – Wang Kah Lun Nov 02 '14 at 05:58
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    There’s nothing wrong with proving them by induction. They can also be proved combinatorially without too much difficulty. For more on the first one see this question and answers. The second is equivalent to the first via the symmetry of the binomial coefficient. – Brian M. Scott Nov 02 '14 at 05:59

3 Answers3

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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

lab bhattacharjee
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$\bf{My\; Solution::}$ 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 $\displaystyle (1+x)^{n+1}-(1+x)^r = \binom{n+1}{r+1}$

juantheron
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$\bf{My\; Solution::}$ 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 $\displaystyle \left\{(1+x)^{r+k+1}-(1+x)^{r}\right\} = \binom{r+k+1}{r+1} = \binom{r+k+1}{k}$

juantheron
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