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

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Solution

$= \frac{n!}{(n - r)! r!} + \frac{n!}{(n - r + 1)! (r - 1)!}$
$= n! [\frac{1}{(n - r)! r!} + \frac{1}{(n - r + 1)! (r - 1)!}]$
$= n! [\frac{1}{(n - r)! r (r - 1)!} + \frac{1}{(n - r + 1)! (r - 1)!}]$
$= \frac{n!}{(r - 1)!} [\frac{1}{(n - r)! r} + \frac{1}{(n - r + 1)!}]$
$= \frac{n!}{(r - 1)!} [\frac{1}{r (n - r)!} + \frac{1}{(n - r + 1) (n - r)!}]$
$= \frac{n!}{(r - 1)! (n - r)!} [\frac{1}{r} + \frac{1}{(n - r + 1)}]$
$= \frac{n!}{(r - 1)! (n - r)!} [\frac{n - r + 1 + r}{r (n - r + 1)}]$
$= \frac{n!}{(r - 1)! (n - r)!} [\frac{n + 1}{r (n - r + 1)}]$
$= \frac{(n + 1) n!}{(n - r + 1) (r - 1)! r (n - r)!}$
$= \frac{(n + 1)!}{(n - r + 1)! r!}$
$=^{n + 1} C_{r}$

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Similar questions

^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}

^{n} C_{r - 1} +^{n} C_{r} =^{n + 1} C_{r}

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

$^{n} C_{r} +^{n} C_{r - 1}$ is equal to

(2 \leq r \leq n),

^{n} C_{r} + 2 \cdot^{n} C_{r + 1} +^{n} C_{r + 2}

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