The expression n C r -2 n C r -1- n C r -2 is equal toA- n -1 C rB- n -2 C r -1C- n -2 C rD- n -1 C r -1

The correct option is C ^n+2C_rGiven nbsp; ^nC_r+2 ^nC_r 1+ ^nC_r 2 nbsp;= nbsp; ^nC_r+ ^nC_r 1+ ^nC_r 1+ ^nC_r 2 nbsp;= ^n+1C_r+ ^n+1C_r 1 nbsp;= ...

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Standard XII

Mathematics

Pascal Triangle

The expressio...

Question

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

^{n + 1} C_{r}

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

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^{n + 2} C_{r}

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^{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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Similar questions

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

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

Q. If

(2 \leq r \leq n),

then

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

is equal to

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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The expression nCr+2 nCr−1+ nCr−2 is equal to

The correct option is C n+2CrGiven nCr+2 nCr−1+ nCr−2= nCr+ nCr−1+ nCr−1+ nCr−2= n+1Cr+ n+1Cr−1= n+2Cr

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

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