The expression $^{n} C_{r} +^{n - 1} C_{r} + . . . . . . . . . +^{r} C_{r}$ is equal to :

$^{n + 1} C_{r}$

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$^{n + 1} C_{r + 1}$

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$^{n + 2} C_{r}$

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$2^{n}$

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Solution

The correct option is B
$^{n + 1} C_{r + 1}$

Explanation For The Correct Option
Evaluating the expression
The Given expression:
$^{n} C_{r} +^{n - 1} C_{r} + . . . . . . . . . +^{r} C_{r}$
$\Rightarrow^{n} C_{r} +^{n - 1} C_{r} +^{n - 2} C_{r} + . . . . . . +^{r + 2} C + (^{r + 1} C_{r} +^{r + 1} C_{r + 1} ) $
We know that $^{n + 1} C_{r + 1} =^{n} C_{r + 1} +^{n} C_{r}$
$∴^{n} C_{r} +^{n - 1} C_{r} n +^{n - 2} C_{r} + . . . . . . +^{r + 2} C_{r} +^{r + 2} C_{r + 1}$
$\Rightarrow^{n} C_{r} +^{n - 1} C_{r} n +^{n - 2} C_{r} + . . . . . . +^{r + 3} C_{r + 1}$
By continuing this, we get
$\Rightarrow^{n} C_{r} + (^{n - 1} C_{r} +^{n - 1} C_{r + 1}) \Rightarrow^{n} C_{r} +^{n} C_{r + 1} \Rightarrow^{n + 1} C_{r + 1}$
Hence, the correct answer is Option (B),

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