- A$2^n.n!$
- B$1\cdot3\cdot5 ........ (2n-1)$
- C$2^n\left\{1\cdot3\cdot5 ........ (2n-1)\right\}$
- ✓$2^n\left\{1\cdot3\cdot5 ........ (2n-1)\right\}n!$
$(2n)! = 1\cdot2\cdot3\cdot4 .... (2n-1) \cdot2n$
$= 2\cdot4 ..... (2n) \cdot1\cdot3\cdot5 ... (2n-1)$
$= (2 \times 1) \cdot (2 \times 2) ... (2 \times n) \cdot 1\cdot3\cdot5 ... (2n-1)$
$= (2 \times 2 \times .... \times n $ વખત) $(1\cdot 3 \cdot 5 \cdot ..............(2n-1)$) $\left\{1\cdot3\cdot 5 .... (2n-1)\right\}$
$= 2^n \cdot\left\{1\cdot3\cdot5.... (2n-1)\right\} n!$
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વિધાન $1:$ $\left| {{Z_1} - {Z_2}} \right|\, \ge \left| {{Z_{_1}}} \right|\, - \,\left| {{Z_{_2}}} \right|$
વિધાન $2:$ $\left| {{Z_1} + {Z_2}} \right|\, \le \left| {{Z_{_1}}} \right|\, + \,\left| {{Z_{_2}}} \right|$
