The greatest positive integer which divides nn-1n-2-n-r-1- n- N is

The correct option is D r! P( n ) =n(n+1)(n+2).......(n+r 1) where, P(r) nbsp;product of r consecutive numbers which is divisible by ...

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Standard XII

Mathematics

Sequence

The greatest ...

Question

The greatest positive integer which divides $n (n + 1) (n + 2) \dots (n + r - 1) \forall n \in N$ is

r!

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(r + 1)!

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(n + r)

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(n - r + 1)

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Solution

The correct option is D $r!$
$P (n) = n (n + 1) (n + 2) . . . . . . . (n + r - 1)$
where, $P (r)$ product of $r$ consecutive numbers which is divisible by $r!$
Ans: A

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Similar questions

Q. Prove that:
(i)

\frac{n!}{(n - r)!}

= n (n − 1) (n − 2) ... (n − (r − 1))

(ii)

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

Prove that:
(i) $\frac{n!}{(n - r)!}$
= n(n-1)(n-2)...(n-(r-1))
(ii) $\frac{n!}{(n - r)! r!} + \frac{n!}{(n - r + 1)! (r - 1)!}$
= $\frac{(n + 1)!}{r! (n - r + 1)!}$

Q. Prove that if

n

and

r

are positive integers

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

is equal to

0

r

be less than

n

, and to

n!

r = n

Q. Prove that:

\frac{n!}{(n - r)!} = n (n - 1) (n - 2) . . . (n - (r - 1))

Q. If

(1 - x)^{- n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . + a_{r} x^{r} + . . . . .

, then

a_{0} + a_{1} + a_{2} + . . . . . . + a_{r}

is equal to

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The greatest positive integer which divides n(n+1)(n+2)…(n+r−1)∀n∈N is

The correct option is D r!P(n)=n(n+1)(n+2).......(n+r−1)where, P(r) product of r consecutive numbers which is divisible by r!Ans: A

The greatest positive integer which divides $n (n + 1) (n + 2) \dots (n + r - 1) \forall n \in N$ is

The correct option is D $r!$
$P (n) = n (n + 1) (n + 2) . . . . . . . (n + r - 1)$
where, $P (r)$ product of $r$ consecutive numbers which is divisible by $r!$
Ans: A