a-n-1-2n-1-a-n-a-1-1-a-n-

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

a_{n + 1} = (2 n + 1) a_{n}

a_{1} = 1

a_{n} = ?

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

a_{n} = (2 n - 1) a_{n - 1} = (2 n - 1) \dots 3 \times 1 = (2 n - 1)!!

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

\frac{a_{n + 1}}{a_{n}} = (2 n + 1) \Rightarrow \prod_{k = 1}^{n - 1} \frac{a_{k + 1}}{a_{k}} = \prod_{k = 1}^{n - 1} (2 k + 1) \Rightarrow

\frac{a_{2}}{a_{1}} \times \frac{a_{3}}{a_{2}} \times \dots . \times \frac{a_{n}}{a_{n - 1}} = 3 \times 5 \times 7 \times \dots . (2 n - 1) \Rightarrow

a_{n} = 2 \times 3 \times 4 \times 5 \times 6 \dots \dots (2 n - 1) \times 2 n \times \frac{1}{2 \times 4 \times 6 \times \dots . (2 n)}

= \frac{(2 n)!}{2^{n} \times n!} \Rightarrow a_{n} = \frac{(2 n)!}{n! \times 2^{n}}

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

a_{1} = 1

a_{2} = 3

\dots

a_{n} = \underset{k = 0}{\prod^{n - 1}} (2 k + 1)

Commented by ckkim89 last updated on 09/May/20

oh, thanks!:)