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Pascal Triangle

The expressio...

Question

The expression $^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2}$ is equal to

A

^{n + 1} C_{r}

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B

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

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C

^{n + 2} C_{r}

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D

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

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Solution

The correct option is C $^{n + 2} C_{r}$
Given
$^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2} =^{n} C_{r} +^{n} C_{r - 1} +^{n} C_{r - 1} +^{n} C_{r - 2} =^{n + 1} C_{r} +^{n + 1} C_{r - 1} =^{n + 2} C_{r}$

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

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

Q. Find the value of

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

.

Let r and n be positive integers such that $1 \leq r \leq n .$ Then prove the following :

(i) $\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$
(ii) $n^{n - 1} C_{r - 1} = (n - r + 1)^{n} C_{r - 1}$
(iii) $\frac{^{n} C_{r}}{^{n - 1} C_{r - 1}} = \frac{n}{r}$
(iv) $^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2} =^{n + 2} C_{r}$

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Pascal Triangle

Standard XII Mathematics

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