The value of -r-0n-1 nCr - nCr-nCr-1 equals

The correct option is A n/2 ∑_r=0^n 1^nC_r^nC_r+^nC_r+1 =∑_r=0^n 111+^nC_r+1^nC_r =∑_r=0^n 111+n rr+1 =∑_r=0^n 1r+1n+1=1n+1∑_r=0^n 1 ...

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Standard XII

Mathematics

Definition of Functions

The value of ...

Question

The value of $n - 1 \sum r = 0^{n} C_{r} / (^{n} C_{r} +^{n} C_{r + 1})$ equals

n + 1

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n / 2

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n + 2

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none of these

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Solution

The correct option is A $n / 2$
$n - 1 \sum r = 0 \frac{^{n} C_{r}}{^{n} C_{r} +^{n} C_{r + 1}}$ $= n - 1 \sum r = 0 \frac{1}{1 + \frac{^{n} C_{r + 1}}{^{n} C_{r}}}$ $= n - 1 \sum r = 0 \frac{1}{1 + \frac{n - r}{r + 1}}$
$= n - 1 \sum r = 0 \frac{r + 1}{n + 1} = \frac{1}{n + 1} n - 1 \sum r = 0 (r + 1)$
$= \frac{1}{(n + 1)} [1 + 2 + . . . . + n] = \frac{n}{2}$

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The value of n−1∑r=0nCr/(nCr+nCr+1) equals

The correct option is A n/2n−1∑r=0nCrnCr+nCr+1 =n−1∑r=011+nCr+1nCr =n−1∑r=011+n−rr+1=n−1∑r=0r+1n+1=1n+1n−1∑r=0(r+1)=1(n+1)[1+2+....+n]=n2

The value of $n - 1 \sum r = 0^{n} C_{r} / (^{n} C_{r} +^{n} C_{r + 1})$ equals