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Question Number 98673 by MJS last updated on 15/Jun/20

find a_{n} in terms of n

(I {can}^{'} t find it \dots)

a_{1} = 1; a_{2} = 4

a_{3} = a_{2} \times 4 \times \frac{2^{2} - 1}{2^{2}}

a_{4} = a_{3} \times 4 \times \frac{2^{2} - 1}{2^{2}} \times \frac{3^{2} - 1}{3^{2}}

a_{5} = a_{4} \times 4 \times \frac{2^{2} - 1}{2^{2}} \times \frac{3^{2} - 1}{3^{2}} \times \frac{4^{2} - 1}{4^{2}}

\dots

n ⩾ 2 : a_{n + 1} = 4 a_{n} \underset{k = 2}{\prod^{n}} \frac{k^{2} - 1}{k^{2}}

Answered by M±th+et+s last updated on 16/Jun/20

s i r w h a t d o y o u t h i n k a b o u t i t

\underset{k = 2}{\sum^{n}} \frac{k^{2} - 1}{k^{2}} = \frac{1}{2 + \frac{a_{n} - 1}{a n + 1}}

Answered by maths mind last updated on 16/Jun/20

\Leftrightarrow \frac{a_{n + 1}}{a_{n}} = 4 . \underset{k = 2}{\prod^{n}} . \frac{(k - 1) (k + 1)}{k . k} .

\underset{k = 2}{\prod^{n}} \frac{k + 1}{k} = \frac{n + 1}{2}

\underset{k = 2}{\prod^{n}} \frac{k - 1}{k} = \frac{1}{n}

\Rightarrow \frac{a_{n + 1}}{a_{n}} = \frac{2 (n + 1)}{n}

\Rightarrow \underset{j = 1}{\prod^{m - 1}} \frac{a_{j + 1}}{a_{j}} = \underset{j = 1}{\prod^{m - 1}} \frac{2 (n + 1)}{n} \Rightarrow \frac{a_{m}}{a_{1}} = 2^{m - 1} . m

\Rightarrow a_{m} = a_{1} . m . 2^{m - 1}

Commented by MJS last updated on 16/Jun/20

great, thank you! I {hadn}^{'} t thought of this

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