Which of the following hold good? If n is a +ve integer, then Options: A . &nbsp n n > 1.3.5 … … ( 2 n − 1 ) B . &nbsp 2.4.6 … … 2 n < ( n + 1 ) n C . &nbsp ( n ! ) 3 < n n ( n + 1 2 ) 2 n D . &nbsp [ 1 r + 2 r + 3 r + … … + n r ] n > n n . ( n ! ) r E . &nbsp b 2 + c 2 b + c + c 2 + a 2 c + a + a 2 + b 2 a + b > a + b + c

Which of the following hold good? If n is a +ve integer, then

Discussion Forum : Inequalities Modulus And Logarithms

Question -

Which of the following hold good? If n is a +ve integer, then

Options:

A .

n^{n} > 1.3.5 \dots \dots (2 n - 1)

B .

2.4.6 \dots \dots 2 n < (n + 1)^{n}

C .

(n!)^{3} < n^{n} {(\frac{n + 1}{2})}^{2 n}

D .

[1^{r} + 2^{r} + 3^{r} + \dots \dots + n^{r}]^{n} > n^{n} . (n!)^{r}

E .

\frac{b^{2} + c^{2}}{b + c} + \frac{c^{2} + a^{2}}{c + a} + \frac{a^{2} + b^{2}}{a + b} > a + b + c

Answer: Option A
:
A, B, C, and D
(a), (b), (c), (d)
(a) Apply A.M. of 1,3,5 .... (2n-1), i.e. of n numbers is > G.M. and remember that 1+3+5+... upto n terms

= \frac{n}{2} [2 a + (n - 1) d] = n^{2}

as a=1 and d=2
(b) Proceed as in part (a)
(c) Apply A.M. of

1^{3}, 2^{3}, \dots \dots n^{3}

is greater than their G.M. and remember that

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

(d)

\frac{1^{r} + 2^{r} + 3^{r} + \dots \dots + n^{r}}{n} > [1^{r} . 2^{r} . 3^{r} . \dots \dots . n^{r}]^{\frac{1}{n}}

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

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