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$$
\text { If }(2 \leq r \leq n), \text { then }{ }^{n} C_{r}+2 \cdot{ }^{n} C_{r+1}+{ }^{n} C_{r+2} \text { is }
$$
equal to
Options:
\text { If }(2 \leq r \leq n), \text { then }{ }^{n} C_{r}+2 \cdot{ }^{n} C_{r+1}+{ }^{n} C_{r+2} \text { is }
$$
equal to
Solution:
1436 Upvotes
Verified Answer
The correct answer is:
${ }^{n+2} C_{r+2}$
${ }^{n} C_{r}+2 \cdot{ }^{n} C_{r+1}+{ }^{n} C_{r+2}$
$={ }^{n} C_{r}+{ }^{n} C_{r+1}+{ }^{n} C_{r+1}+{ }^{n} C_{r+2}$
$={ }^{n+1} C_{r+1}+{ }^{n+1} C_{r+2}$
$\quad\left(\because{ }^{n} C_{r}+{ }^{n} C_{r+1}={ }^{n+1} C_{r+1}\right)$
$={ }^{n+2} C_{r+2}$
$={ }^{n} C_{r}+{ }^{n} C_{r+1}+{ }^{n} C_{r+1}+{ }^{n} C_{r+2}$
$={ }^{n+1} C_{r+1}+{ }^{n+1} C_{r+2}$
$\quad\left(\because{ }^{n} C_{r}+{ }^{n} C_{r+1}={ }^{n+1} C_{r+1}\right)$
$={ }^{n+2} C_{r+2}$
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