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

The correct option is C ( n+2 r )( 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 ...

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Number of Terms in Binomial Expansion

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

lim x \to \infty \frac{1 \cdot n \sum r = 1 r + 2 \cdot n - 1 \sum r = 1 r + 3 \cdot n - 2 \sum r = 1 r + . . . . + n \cdot 1}{n^{4}}

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n^{r} - n {(n - 1)}^{r} + \frac{n (n - 1)}{2!} {(n - 2)}^{r} - \frac{n (n - 1) (n - 2)}{3!} {(n - 3)}^{r} + \dots

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lim n \to \infty \frac{n \sum r = 1 r + 2. n - 1 \sum r = 1 r + 3. n - 2 \sum r = 1 r + . . . . . . + n .1}{n^{4}}

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})

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