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Question Number 112162 by 175mohamed last updated on 06/Sep/20

Commented by JDamian last updated on 06/Sep/20

$$Please,definen!!$$

Commented by mnjuly1970 last updated on 06/Sep/20

$$n!!\left(doublefactorial\right)$$$$ifn=2k\Rightarrow$$$$\left(2k\right)!!=(2k-2)\ast (2k-4)\ast (2k-6)....\ast 2$$$$ifn=2k-1\Rightarrow$$$$(2k-1)\ast (2k-3)\ast (2k-5).....\ast 3\ast 1$$$$forexample$$$$8!!=8\ast 6\ast 4\ast 2=??$$$$9!!=9\ast 7\ast 5\ast 3\ast 1$$

Commented by MJS_new last updated on 06/Sep/20

$$n!!\{\begin{array}{l}n=2k;\left(2k\right)!!=2^k\times k!\\ n=2k+1;(2k+1)!!=\frac{\left(2k\right)!}{2^k\times k!}\end{array}$$

Answered by mnjuly1970 last updated on 06/Sep/20

$$=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{(2k-1)(2k-3)(2k-5)..\ast 3\ast 1}{k!\ast 3^k}$$$$=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{\left(2k\right)(2k-1)(2k-2)(2k-3)(2k-4)(2k-5)....\left(3\right)2\left(1\right)}{2^k\ast k!\ast k!\ast 3^k}$$$$\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{6^k}\left({}_k^{2k}\right)=\frac{1}{3}+\frac{1}{6}+....=A$$$$A=-1+(1+\frac{1}{3}+\frac{1}{6}+...)....\left(1\right)$$$$ontheotherhandweknow:$$$${(1+x)}^n=1+nx+\frac{n(n-1)}{2}{x}^2+...\left(2\right)$$$$from\left(1\right),\left(2\right):nx=\frac{1}{3},\frac{n(n-1)}{2}{x}^2=\frac{1}{6}$$$$\frac{{\left(nx\right)}^2}{2}-\frac{x\left(nx\right)}{2}=\frac{1}{6}\Rightarrow \frac{1}{18}-\frac{x}{6}=\frac{1}{6}$$$$\frac{1}{18}-\frac{1}{6}=\frac{x}{6}\Rightarrow x=-\frac{2}{3}$$$$n=\frac{-1}{2}$$$$\mathbf{÷}{(1-\frac{2}{3})}^{-\frac{1}{2}}=\sqrt{3}\Rightarrow A=-1+\sqrt{3}✓✓✓$$$$$$$$$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$$$$$$$$$$$

Commented by mohammad17 last updated on 07/Sep/20

$$sirbatthequestionfrom(n=0to\mathrm{\infty })how$$$$becamefrom(n=1to\mathrm{\infty })?$$

Commented by mnjuly1970 last updated on 07/Sep/20

$$(-1)!!=!!!$$$$$$

Commented by MJS_new last updated on 07/Sep/20

$$(-1)!!\mathrm{and}0!!\mathrm{are}\mathrm{defined}=1$$

Commented by mohammad17 last updated on 07/Sep/20

$$thankyousir$$

Commented by mohammad17 last updated on 07/Sep/20

$$thankyousir$$

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