If 2 - r - n - then n C r -2 - n C r -1- n C r -2 is equal toA- 2 - n C r -2B- n -1 C r -1C - n -2 C r -2D- n -1 C r

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Standard XII

Mathematics

Integration

If 2 ≤ r ≤ n ...

Question

If $(2 \leq r \leq n),$ then $^{n} C_{r} + 2 \cdot^{n} C_{r + 1} +^{n} C_{r + 2}$ is equal to

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

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

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

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^{n + 1} C_{r}

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Solution

The correct option is C $^{n + 2} C_{r + 2}$
Here, we can write the above expression as:
$^{n} C_{r} +^{n} C_{r + 1} +^{n} C_{r + 1} +^{n} C_{r + 2}$
$=^{n + 1} C_{r + 1} +^{n + 1} C_{r + 2}$
$=^{n + 2} C_{r + 2}$
[Using the property $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ ]

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Similar questions

Q. The expression

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

is equal to

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

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

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. Prove that

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

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If (2≤r≤n), then nCr+2⋅nCr+1+nCr+2 is equal to

The correct option is C n+2Cr+2Here, we can write the above expression as: nCr+nCr+1+nCr+1+nCr+2 = n+1Cr+1+n+1Cr+2 = n+2Cr+2 [Using the property nCr+nCr−1= n+1Cr]

If $(2 \leq r \leq n),$ then $^{n} C_{r} + 2 \cdot^{n} C_{r + 1} +^{n} C_{r + 2}$ is equal to

The correct option is C $^{n + 2} C_{r + 2}$
Here, we can write the above expression as:
$^{n} C_{r} +^{n} C_{r + 1} +^{n} C_{r + 1} +^{n} C_{r + 2}$
$=^{n + 1} C_{r + 1} +^{n + 1} C_{r + 2}$
$=^{n + 2} C_{r + 2}$
[Using the property $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ ]