Reduction formula q (1 Viewer)

=)(= · Nov 27, 2021

part c please

5uckerberg · Nov 27, 2021

Let me give you some clues note the numerator is a multiple of even numbers and the denominator is giving us odd numbers going towards 2n+1 ending with 2n+3. If you had done part ii this should be clear.

Next you want (2n+3)! how do you get there?
Once you know how then this is a cakewalk. Try the last question from the 2020 HSC that Sydney Morning Herald our on the front cover of the newsletter. Should be good practice for this question. If you can do this question Jordan Baker should broadcast you to show everyone how it is done.

Trebla · Nov 27, 2021

$T_{k+1} \\= \dfrac{2(k+1)}{2(k+1)+3}T_k\\=\dfrac{2(k+1)}{2(k+1)+3}\times\dfrac{k!(k+1)!}{(2k+3)!}4^{k+1}\\=\dfrac{2(k+1)}{2(k+1)+3}\times\dfrac{2(k+2)}{2(k+2)}\times\dfrac{k!(k+1)!}{(2k+3)!}4^{k+1}$

...and the rest follows

Reduction formula q (1 Viewer)

=)(=

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