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

^{n + 1} C_{r}

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^{n} C_{r + 1}

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

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^{n + 2} C_{r}

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Solution

The correct option is C $^{n + 2} C_{r}$
$\Rightarrow^{n} C_{r} +^{n} C_{r - 1} +^{n} C_{r - 1} +^{n} C_{r - 2} . . . . . . . (1) \Rightarrow^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$
$(1)$ can be written as
$\Rightarrow^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r} \Rightarrow^{n + 1} C_{r} +^{n + 1} C_{r - 1} =^{n + 2} C_{r}$
Hence the correct answer is $^{n + 2} C_{r}$

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Similar questions

$^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C r - 2$ =

Q. The expression

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

is equal to

Q. If

(2 \leq r \leq n),

then

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

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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) $^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2} =^{n + 2} C_{r}$

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

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nCr+2 nCr−1+nCr−2 is equal to

The correct option is C n+2Cr⇒nCr+nCr−1+nCr−1+nCr−2.......(1)⇒nCr+nCr−1=n+1Cr(1) can be written as ⇒nCr+nCr−1=n+1Cr⇒n+1Cr+n+1Cr−1=n+2CrHence the correct answer is n+2Cr

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