Prove that $^{n} C_{r} +^{n - 1} C_{r} +^{n - 2} C_{r} + . . . . . +^{r} C_{r} =^{n + 1} C_{r + 1}$ .

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Solution

To prove:
$^{n} C_{r} +^{n - 1} C_{r} +^{n - 2} C_{r} + . . . . . . +^{r} C_{r} =^{n + 1} C_{r + 1}$
$^{n} C_{r} +^{n - 1} C_{r} +^{n - 2} C_{r} + . . . . . +^{r + 1} C_{r} +^{r} C_{r} =^{n + 1} C_{r + 1}$
$[∵^{n} C_{n} =^{n + 1} C_{n + 1}]$
$\Rightarrow^{n} C_{r} +^{n - 1} C_{r} +^{n - 2} C_{r} + . . . . . . +^{r + 2} C_{r} + (^{r + 1} C_{r} +^{r + 1} C_{r + 1})$
$\Rightarrow^{r + 2} C_{r + 1}$
[We know that $^{n + 1} C_{r + 1} =^{n} C_{r + 1} +^{n} C_{r}$ ]
$∴^{n} C_{r} +^{n - 1} C_{r} +^{n - 2} C_{r} + . . . . . . +^{r + 2} C_{r} +^{r + 2} C_{r + 1} =^{n + 1} C_{r + 1}$
$\Rightarrow^{n} C_{r} +^{n} - 1 C_{r} +^{n - 2} C_{r} + . . . . . . +^{r + 3} C_{r + 1} =^{n + 1} C_{r + 1}$
By continuing this, we get
$\Rightarrow^{n} C_{r} + (^{n - 1} C_{r} +^{n - 1} C_{r + 1}) =^{n + 1} C_{r + 1}$
$\Rightarrow^{n} C_{r} +^{n} C_{r + 1} =^{n + 1} C_{r + 1}$
$\Rightarrow^{n + 1} C_{r + 1} =^{n + 1} C_{r + 1}$
L.H.S $=$ R.H.S
[Hence proved].

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Q. Find the value of :

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Prove that $^{n} C_{r} +^{n - 1} C_{r} +^{n - 2} C_{r} + . . . . . +^{r} C_{r} =^{n + 1} C_{r + 1}$ .