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Question Number 39519 by math khazana by abdo last updated on 07/Jul/18

s i m p l i f y

1) A_{n} = \frac{1}{\sqrt{a}} {{(\frac{1 + \sqrt{a}}{2})}^{n} - {(\frac{1 - \sqrt{a}}{2})}^{n}} w i t h n n a t u r a l

i n t e g r a n d a > 0

2) f (x) = \frac{1}{\sqrt{2 x + 1}} {{(\frac{1 + \sqrt{2 x + 1}}{2})}^{n} - {(\frac{1 - \sqrt{2 x + 1}}{2})}^{n}}

Commented by math khazana by abdo last updated on 08/Jul/18

1) w e h a v e A_{n} = \frac{1}{2^{n} \sqrt{a}} {{(1 + \sqrt{a})}^{n} - {(1 - \sqrt{a})}^{n}}

= \frac{1}{2^{n} \sqrt{a}} {\sum_{k = 0}^{n} C_{n}^{k} {(\sqrt{a})}^{k} - \sum_{k = 0}^{n} C_{n}^{k} {(- \sqrt{a})}^{k}}

= \frac{1}{2^{n} \sqrt{a}} {\sum_{k = 0}^{n} C_{n}^{k} ({(\sqrt{a})}^{k} - {(- \sqrt{a})}^{k})}

= \frac{1}{2^{n} \sqrt{a}} {\sum_{k = 0}^{n} C_{n}^{k} (1 - {(- 1)}^{k}) {(\sqrt{a})}^{k}}

= \frac{1}{2^{n} \sqrt{a}} \sum_{p = 0}^{[\frac{n - 1}{2}]} 2 C_{n}^{2 p + 1} {(\sqrt{a})}^{2 p + 1}}

A_{n} = \frac{1}{2^{n - 1}} \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} a^{p}

2) l e t t a k e a = 2 x + 1 \Rightarrow

f (x) = \frac{1}{2^{n - 1}} \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} {(2 x + 1)}^{p} .

Answered by tanmay.chaudhury50@gmail.com last updated on 08/Jul/18

1) A_{n} = \frac{1}{2^{n} \sqrt{a}} {(1 + n_{C_{1}} (\sqrt{a}) + n_{C_{2}} {(\sqrt{a})}^{2} + n_{C_{3}} {(\sqrt{a})}^{3} +

+ \dots + {(\sqrt{a})}^{n}) - (1 - n_{C_{1}} {(\sqrt{a})}^{2} + n_{C_{2}} {(\sqrt{a})}^{3} - \dots)}

= \frac{2}{2^{n} \sqrt{a}} {n_{C_{1}} (\sqrt{a}) + n_{C_{3}} {(\sqrt{a})}^{3} + n_{C_{5}} {(\sqrt{a})}^{5} + . .