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

( n r )+2 ( n r 1 )+ ( n r 2 )= [ ( n r )+ ( n r 1 ) ] [ ( n r 1 )+( n r 2 ) ]=( n+1 r )+( n+1 r 1 )=( n+2 r ) [∵ ^nC_r + ^nC_r ...

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Binomial Coefficients

For 2 ≤ r ≤ n...

Question

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

(\begin{matrix} n + 1 r - 1 \end{matrix})

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(\begin{matrix} n + 1 r + 1 \end{matrix})

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2 (\begin{matrix} n + 2 r \end{matrix})

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(\begin{matrix} n + 2 r \end{matrix})

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Solution

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Similar questions

2 \leq r \leq n

(\begin{matrix} n r \end{matrix}) + 2 (\begin{matrix} n r - 1 \end{matrix}) + (\begin{matrix} n r - 2 \end{matrix}) =

2 \geq r \geq n, (_{n}^{r}) + 2 (_{n}^{r - 1}) + (_{n}^{r - 2})

2 \leq r \leq n, (_{r}^{n + 1}) + (_{r - 1}^{n}) + (_{r - 2}^{n})

n

n^{r} - 1

\frac{n^{r} - n r + r - 1}{n - 1}

lim n \to \infty n \sum r = 1 \frac{r}{n^{2} + n r}

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