The value of - r -044 C r -4 C r -4 C r -1 isA- 1-2- 3- 1-3D- 2

The correct option is B 3∑_r = 0^4 ^4C_r ^4C_r+ ^4C_r+1 nbsp;= ∑_r = 0^4 ^4C_r ^5C_r+1 = ∑_r = 0^4r+15 (∵ ^nC_r+ ^nC_r 1= ^n+1C_r) nbsp;= 15 nbsp;∑_r ...

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Standard XII

Mathematics

Pascal Triangle

The value of ...

Question

The value of $4 \sum r = 0 \frac{^{4} C_{r}}{^{4} C_{r} +^{4} C_{r + 1}}$ is

\frac{1}{2}

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3

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\frac{1}{3}

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2

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Solution

The correct option is B $3$
$4 \sum r = 0 \frac{^{4} C_{r}}{^{4} C_{r} +^{4} C_{r + 1}} = 4 \sum r = 0 \frac{^{4} C_{r}}{^{5} C_{r + 1}} = 4 \sum r = 0 \frac{r + 1}{5} (∵^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}) = \frac{1}{5} 4 \sum r = 0 r + 1 = \frac{1}{5} (1 + 2 + 3 + 4 + 5) = 3$

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Pascal Triangle

Standard XII Mathematics

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The value of 4∑r=0 4Cr 4Cr+ 4Cr+1 is

The correct option is B 34∑r=0 4Cr 4Cr+ 4Cr+1=4∑r=0 4Cr 5Cr+1=4∑r=0r+15(∵ nCr+ nCr−1= n+1Cr)=154∑r=0r+1=15(1+2+3+4+5)=3

The value of $4 \sum r = 0 \frac{^{4} C_{r}}{^{4} C_{r} +^{4} C_{r + 1}}$ is

The correct option is B $3$
$4 \sum r = 0 \frac{^{4} C_{r}}{^{4} C_{r} +^{4} C_{r + 1}} = 4 \sum r = 0 \frac{^{4} C_{r}}{^{5} C_{r + 1}} = 4 \sum r = 0 \frac{r + 1}{5} (∵^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}) = \frac{1}{5} 4 \sum r = 0 r + 1 = \frac{1}{5} (1 + 2 + 3 + 4 + 5) = 3$