LetI n - n n - J n -1-3 -5 - -2 n -1 for all n -1- n - N- thenA- I n - J n B- I n - J n C- I n - J n D- I n -1-2 J n

The correct option is A I(n) gt; J(n)I(n)J(n)= n· n· n· n⋯ n1· 3· 5· n⋯ 2n 1 or =n·2· n·4· n·6· n⋯ 2n· n1· 2· 3· n⋯ 2n =2^n nbsp; n!· n^n (2n)! =2^n· ...

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Byju's Answer

Standard XII

Mathematics

Permutations and Combinations

LetI n = n n ...

Question

Let $I (n) = n^{n}, J (n) = 1.3.5.... (2 n - 1) for all (n > 1), n \in N,$ then

I (n) > J (n)

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I (n) < J (n)

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I (n) \neq J (n)

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I (n) = \frac{1}{2} J (n)

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Solution

The correct option is A $I (n) > J (n)$
$\frac{I (n)}{J (n)} = \frac{n \cdot n \cdot n \cdot n \dots n}{1 \cdot 3 \cdot 5 \cdot n \dots 2 n - 1} or = \frac{n \cdot 2 \cdot n \cdot 4 \cdot n \cdot 6 \cdot n \dots 2 n \cdot n}{1 \cdot 2 \cdot 3 \cdot n \dots 2 n} = \frac{2^{n} n! \cdot n^{n}}{(2 n)!} = \frac{2^{n} \cdot n^{n}}{^{2 n} P_{n}} > 1$

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Similar questions

Q. Let

a > b > 0

and

I (n) = a^{1 / n} - b^{1 / n},

J (n) = (a - b)^{1 / n}

for all

n \geq 2

. then

Q. Let

I_{n} = \int_{0}^{\frac{π}{2}} \frac{s i n (2 n - 1) x}{s i n x} d x, J_{n} = \int_{0}^{\frac{π}{2}} \frac{s i n^{2} n x}{s i n^{2} x} d x, n \in N

, then

Q. The middle term in the expansion of

(1 + x)^{2 n}

The middle term in the expansion of $(1 + x)^{2 n}$ is

Q. If

n

is a positive integer, show that
(1)

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

;
(2)

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

;
the series in each case being extended to

n

terms; and
(3)

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

;
(4)

{(n + p)}^{n} - n {(n + p - 1)}^{n} + \frac{n (n - 1)}{2!} {(n + p - 2)}^{n} - \dots = n!

;
the series in the last two cases being extended to

n + 1

terms.

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LetI(n)=nn,J(n)=1.3.5....(2n−1) for all (n>1),n∈N, then

The correct option is A I(n)>J(n)I(n)J(n)=n⋅n⋅n⋅n⋯n1⋅3⋅5⋅n⋯2n−1or=n⋅2⋅n⋅4⋅n⋅6⋅n⋯2n⋅n1⋅2⋅3⋅n⋯2n=2nn!⋅nn(2n)!=2n⋅nn2nPn>1

Let $I (n) = n^{n}, J (n) = 1.3.5.... (2 n - 1) for all (n > 1), n \in N,$ then