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

^{n + 1} C_{r}

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^{n + 2} C_{r}

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

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

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Solution

The correct option is D $^{n + 2} C_{r + 1}$
$^{n} C_{r + 1} + 2^{n} C_{r} +^{n} C_{r - 1}$
$= (^{n} C_{r + 1} +^{n} C_{r}) + (^{n} C_{r} +^{n} C_{r - 1})$
$=^{n + 1} C_{r + 1} +^{n + 1} C_{r} =^{n + 2} C_{r + 1}$

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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. Prove that

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

Q. If

(2 \leq r \leq n),

then

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

is equal to

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nCr+1+2nCr+nCr−1=

The correct option is D n+2Cr+1nCr+1+2nCr+nCr−1=(nCr+1+nCr)+(nCr+nCr−1)=n+1Cr+1+n+1Cr=n+2Cr+1

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

The correct option is D $^{n + 2} C_{r + 1}$
$^{n} C_{r + 1} + 2^{n} C_{r} +^{n} C_{r - 1}$
$= (^{n} C_{r + 1} +^{n} C_{r}) + (^{n} C_{r} +^{n} C_{r - 1})$
$=^{n + 1} C_{r + 1} +^{n + 1} C_{r} =^{n + 2} C_{r + 1}$