The value of -r-1nrn Cr-n Cr-1 isA- n-12-2B- n2-2C- nn-1-2D- nn-1-2

The correct option is C n(n+1)2∑_r=1^nr ^nC_r ^nC_r 1 nbsp;= ∑_r=1^nr(n r+1)r nbsp;= ∑_r=1^n(n r+1) nbsp;= nbsp;∑_r=1^n(n+1) nbsp;∑_r=1^nr nb ...

The value of ...

Question

The value of $n \sum r = 1 \frac{r^{n} C_{r}}{^{n} C_{r - 1}}$ is

\frac{n (n + 1)}{2}

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\frac{(n + 1)^{2}}{2}

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\frac{n^{2}}{2}

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\frac{n (n - 1)}{2}

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Solution

The correct option is C $\frac{n (n + 1)}{2}$
$n \sum r = 1 \frac{r^{n} C_{r}}{^{n} C_{r - 1}} = n \sum r = 1 \frac{r (n - r + 1)}{r} = n \sum r = 1 (n - r + 1) = n \sum r = 1 (n + 1) - n \sum r = 1 r = (n + 1) n \sum r = 1 1 - n \sum r = 1 r = (n + 1) n - \frac{n (n + 1)}{2} = \frac{n (n + 1)}{2}$

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Pascal Triangle

Standard XII Mathematics

The value of n∑r=1r nCr nCr−1 is

The correct option is C n(n+1)2n∑r=1r nCr nCr−1=n∑r=1r(n−r+1)r=n∑r=1(n−r+1)=n∑r=1(n+1)−n∑r=1r=(n+1)n∑r=11−n∑r=1r=(n+1)n−n(n+1)2=n(n+1)2

The value of $n \sum r = 1 \frac{r^{n} C_{r}}{^{n} C_{r - 1}}$ is

The correct option is C $\frac{n (n + 1)}{2}$
$n \sum r = 1 \frac{r^{n} C_{r}}{^{n} C_{r - 1}} = n \sum r = 1 \frac{r (n - r + 1)}{r} = n \sum r = 1 (n - r + 1) = n \sum r = 1 (n + 1) - n \sum r = 1 r = (n + 1) n \sum r = 1 1 - n \sum r = 1 r = (n + 1) n - \frac{n (n + 1)}{2} = \frac{n (n + 1)}{2}$