Show that $| 1 - | 3 - | 5 - . . . . | 2 n - 1 - ----- - > {(| n - -)}^{n}$ .

Open in App

Solution

$\frac{| r - - | 2 n - r - ----- -}{| n - -} > | n - -$

$\frac{| 1 - - | 2 n - 1 - ------ -}{| n - -} > | n - -$

$\frac{| 2 - - | 2 n - 2 - ------ -}{| n - -} > | n - -$

$\frac{| 3 - - | 2 n - 3 - ------ -}{| n - -} > | n - -$
.
.
.
$\frac{| 2 n - 1 - ------ - | 1 - -}{| n - -} > | n - -$

Multiplying the above equation we get .

$\frac{{(| 1 - - | 2 - - | 3 - - . . . | 2 n - 1 - ------ -)}^{2}}{{| n - -}^{n}} > {| n - -}^{n}$

${(| 1 - - | 2 - - | 3 - - . . . | 2 n - 1 - ------ -)}^{2} > {| n - -}^{2 n}$

$(| 1 - - | 2 - - | 3 - - . . . | 2 n - 1 - ------ -) > {| n - -}^{n}$

Suggest Corrections

Similar questions

\frac{2}{1 -} \frac{3}{5 -} \frac{8}{7 -} \dots \frac{n^{2} - 1}{2 n + 1} = \frac{n (n + 3)}{2}

6

n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot 3 + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot 3^{2} - . . . . .,

n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot \frac{1}{3} + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot \frac{1}{3^{2}} - . . . . .,

n^{n} > 1.3.5.... (2 n - 1)

n ϵ N

n^{n} > 1 \times 3 \times 5 \times 7 \times . . . . . . . \times (2 n - 1)

3

n^{3} + \frac{n (n - 1)}{⌊ 2} (n - 2)^{3} + \frac{n (n - 1) (n - 2) (n - 3)}{⌊ 4} (n - 4)^{3} + . . . . = n^{2} (n + 3) 2^{n - 4}

Join BYJU'S Learning Program