For 2- r- n-nr-2nr-1-nr-2 is equal to

The correct option is D n+2r ^nC_r+ ^nC_r 1+ ^nC_r 1+ ^nC_r 2 =( ^nC_r+ ^nC_r 1)+( ^nC_r 1+ ^nC_r 2) = ^n+1C_r+ ^n+1C_r 1 = ^n+2C_r Th ...

For 2≤ r≤ n...

Question

For $2 \leq r \leq n, (\frac{n}{r}) + 2 (\frac{n}{r - 1}) + (\frac{n}{r - 2})$ is equal to

(\frac{n + 1}{r - 1})

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

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

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

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Solution

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Similar questions

2 \leq r \leq n, (\frac{n}{r}) + 2 (\frac{n}{r - 1}) + (\frac{n}{r + 2}) =

^{n} C_{r - 1} +^{n + 1} C_{r - 1} +^{n + 2} C_{r - 1} + . . . . . . . +^{2 n} C_{r - 1}, =^{2 n + 1} C_{r^{2} - 132} -^{n} C_{r}

r

n \geq r - 1

{^{n} C}_{r - 1} + {^{n + 1} C}_{r - 1} + {^{n + 2} C}_{r - 1} + . . . . + {^{2 n} C}_{r - 1} = {^{2 n + 1} C}_{r^{2} - 182} - {^{n} C}_{r}

n

\sum_{r = 0}^{n} r (^{n} C_{r})^{2} = n (^{2 n - 1} C_{n - 1})

\sum_{r = 0}^{n} (^{n} C_{r})^{2} =^{2 n} C_{n}