Given a_(n+1) = ((2a_n )/((2n+1)(2n+2))) find a


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Question Number 120044 by john santu last updated on 28/Oct/20

$G i v e n a_{n + 1} = \frac{2 a_{n}}{(2 n + 1) (2 n + 2)}$ $f i n d a_{n} .$
	Answered by Olaf last updated on 29/Oct/20

	$Let v_{k} = \frac{a_{k + 1}}{a_{k}} = \frac{2}{(2 k + 1) (2 k + 2)}$ $\underset{k = 0}{\prod^{n}} v_{k} = \underset{k = 0}{\prod^{n}} \frac{a_{k + 1}}{a_{k}} = \frac{a_{n + 1}}{a_{0}}$ $(telescopic product)$ $and \underset{k = 0}{\prod^{n}} v_{k} = \underset{k = 0}{\prod^{n}} \frac{2}{(2 k + 1) (2 k + 2)}$ $\underset{k = 0}{\prod^{n}} v_{k} = \frac{2^{n + 1}}{(1.2) (3.4) (5.6) . . . (2 n + 1) (2 n + 2)}$ $\underset{k = 0}{\prod^{n}} v_{k} = \frac{2^{n + 1}}{(2 n + 2)!}$ $\Rightarrow a_{n + 1} = a_{0} \frac{2^{n + 1}}{(2 n + 2)!}$ $and a_{n} = a_{0} \frac{2^{n}}{(2 n)!}$
	Commented by bramlexs22 last updated on 29/Oct/20

	$w h a t t h e v a l u e o f a_{0} s i r ?$
	Commented by Olaf last updated on 29/Oct/20

	${It}^{'} s impossible to know .$ $It is not given in the problem .$ $a_{0} is simply the first term in the$ $sequence .$
	Answered by Bird last updated on 29/Oct/20

	$\frac{a_{n + 1}}{a_{n}} = \frac{1}{(2 n + 1) (2 n + 2)} \Rightarrow$ $\prod_{k = 0}^{n - 1} \frac{a_{k + 1}}{a_{k}} = \prod_{k = 0}^{n - 1} \frac{1}{(2 k + 1) (2 k + 2)} \Rightarrow$ $\frac{a_{1}}{a_{0}} \times \frac{a_{2}}{a_{1}} \times . . . . \times \frac{a_{n}}{a_{n - 1}} = \frac{1}{\prod_{k = 0}^{n - 1} (2 k + 1) \prod_{k = 0}^{n - 1} (2 k + 2)}$ $\Rightarrow a_{n} = a_{0} \times \frac{1}{\prod_{k = 0}^{n - 1} (2 k + 1) \prod_{k = 0}^{n - 1} (2 k + 2)}$ $b u t \prod_{k = 0}^{n - 1} (2 k + 1) = 1.3.5 . . . (2 n - 1)$ $= 1.2.3.4.5 . . . . . (2 n - 1) . 2 n \times \frac{1}{2.4 . . . (2 n)}$ $= \frac{(2 n)!}{2^{n} n!} a l s o$ $\prod_{k = 0}^{n - 1} (2 k + 2) = \prod_{k = 1}^{n} (2 k)$ $= 2^{n} n! \Rightarrow$ $a_{n} = a_{0} . \frac{2^{n} n!}{(2 n)!} . \frac{1}{2^{n} n!} \Rightarrow a_{n} = \frac{a_{o}}{(2 n)!}$
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