Given-a-n-1-2a-n-2n-1-2n-2-find-a-n-

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) \dots (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 \dots . \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 \dots (2 n - 1)

= 1.2.3.4.5 \dots . . (2 n - 1) . 2 n \times \frac{1}{2.4 \dots (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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