For 2 - r - n - nr - 2 nr - 1 - nr - 2

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Byju's Answer

Standard XII

Mathematics

Combination

For 2 ≥ r ≥...

Question

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

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

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

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

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(_{n + 1}^{r})

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Solution

The correct option is D $(_{n + 1}^{r})$
$^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2}$
$= (^{n} C_{r} +^{n} C_{r - 1}) + (^{n} C_{r - 1} +^{n} C_{r - 2})$
$=^{n + 1} C_{r} +^{n + 1} C_{r - 1} =^{n + 2} C_{r} \Rightarrow (d)$

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Q. For

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

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Which of the following best describes the terms of the series? $\frac{1}{2}$ $^{n} C_{1}$ - $\frac{2}{3}$ $^{n} C_{2}$ + $\frac{3}{4}$ $^{n} C_{3}$ ................... $(- 1)^{n + 1}$ $\frac{1}{n + 1}$ $^{n} C_{n}$

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For 2≥r≥n,(rn)+2(r−1n)+(r−2n)

The correct option is D (rn+1)nCr+2nCr−1+nCr−2=(nCr+nCr−1)+(nCr−1+nCr−2)=n+1Cr+n+1Cr−1=n+2Cr⇒(d)

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

The correct option is D $(_{n + 1}^{r})$
$^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2}$
$= (^{n} C_{r} +^{n} C_{r - 1}) + (^{n} C_{r - 1} +^{n} C_{r - 2})$
$=^{n + 1} C_{r} +^{n + 1} C_{r - 1} =^{n + 2} C_{r} \Rightarrow (d)$