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Question Number 92777 by ckkim89 last updated on 09/May/20

$${\mathrm{a}}_{\mathrm{n}+1}=(2\mathrm{n}+1){\mathrm{a}}_{\mathrm{n}}$$$${\mathrm{a}}_1=1$$$${\mathrm{a}}_{\mathrm{n}}=?$$$$$$

Commented by mr W last updated on 09/May/20

$$a_n=(2n-1)a_{n-1}=(2n-1)...3\times 1=(2n-1)!!$$

Commented by mathmax by abdo last updated on 09/May/20

$$\frac{a_{n+1}}{a_n}=(2n+1)\Rightarrow \prod _{k=1}^{n-1}\frac{a_{k+1}}{a_k}=\prod _{k=1}^{n-1}(2k+1)\Rightarrow$$$$\frac{a_2}{a_1}\times \frac{a_3}{a_2}\times ....\times \frac{a_n}{a_{n-1}}=3\times 5\times 7\times ....(2n-1)\Rightarrow$$$$a_n=2\times 3\times 4\times 5\times 6......(2n-1)\times 2n\times \frac{1}{2\times 4\times 6\times ....\left(2n\right)}$$$$=\frac{\left(2n\right)!}{2^n\times n!}\Rightarrow a_n=\frac{\left(2n\right)!}{n!\times 2^n}$$

Answered by prakash jain last updated on 09/May/20

$$a_1=1$$$$a_2=3$$$$...$$$$a_n=\underset{k=0}{\prod ^{n-1}}(2k+1)$$

Commented by ckkim89 last updated on 09/May/20

oh, thanks!:)

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