if a>1 , show ((Σ_(k=1) ^(a^2 −1) (√(a+(√k))))/(Σ_(k=1) ^(a^2 −1) (√(a−(√k))))) = (√2) + 1


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Question Number 192009 by universe last updated on 05/May/23

$i f a > 1, s h o w$ $\frac{\underset{k = 1}{\sum^{a^{2} - 1}} \sqrt{a + \sqrt{k}}}{\underset{k = 1}{\sum^{a^{2} - 1}} \sqrt{a - \sqrt{k}}} = \sqrt{2} + 1$
	Answered by Skabetix last updated on 05/May/23

	Commented bySkabetix last updated on 05/May/23

	$S_{2} = S_{1} (\sqrt{} 2 - 1)$ $\to S_{1} = \frac{S_{2}}{\sqrt{2} - 1}$ $\to \frac{S_{1}}{S_{2}} = \frac{\frac{S_{2}}{\sqrt{2} - 1}}{\frac{S_{2}}{1}} = \frac{S_{2}}{\sqrt{2} - 1} \times \frac{1}{S_{2}} = \frac{1}{\sqrt{2} - 1} = \frac{\sqrt{2} + 1}{(\sqrt{} 2 - 1) (\sqrt{} 2 + 1)} = \sqrt{} 2 + 1$
	Answered by York12 last updated on 07/May/23
	$let Σ_(k=1 ) ^(a^2 −1) (√(a+(√k) ))=s_1 and Σ_(k=1) ^(a^2 −1) (√(a−(√k)))=s_2 s_1 −s_2 =Σ_(k=1) ^(a^2 −1) [(√(((√(a+(√k)))−(√(a−(√k))))^2 ))]=Σ_(k=1) ^(a^2 −1) (√2)(√(a−(√(a^2 −k)))) → [I] Now since Σ_(k=1) ^(a^2 −1) T_k =Σ_(k=1) ^(a^2 −1) T_([(a^2 −1)−(k−1)]) =Σ_(k=1) ^(a^2 −1) T_((a^2 −k)) let T_k =(√(a−(√k) )) → T_((a^2 −k)) =(√(a−(√(a^2 −k)))) ∴ I = Σ_(k=1) ^(a^2 −1) (√(a−(√k)))=(√2)s_2 ∴ s_1 +s_2 =(√2)s_2 → s_1 =(1+(√2))s_2 → (s_1 /s_2 )=(1+(√2)) → (That′s it ) {BY YORK} Telegram : bengubler$
	$l e t \underset{k = 1}{\sum^{a^{2} - 1}} \sqrt{a + \sqrt{k}} = s_{1} a n d \underset{k = 1}{\sum^{a^{2} - 1}} \sqrt{a - \sqrt{k}} = s_{2}$ $s_{1} - s_{2} = \underset{k = 1}{\sum^{a^{2} - 1}} [\sqrt{{(\sqrt{a + \sqrt{k}} - \sqrt{a - \sqrt{k}})}^{2}}] = \underset{k = 1}{\sum^{a^{2} - 1}} \sqrt{2} \sqrt{a - \sqrt{a^{2} - k}} \to [I]$ $N o w s i n c e \underset{k = 1}{\sum^{a^{2} - 1}} T_{k} = \underset{k = 1}{\sum^{a^{2} - 1}} T_{[(a^{2} - 1) - (k - 1)]} = \underset{k = 1}{\sum^{a^{2} - 1}} T_{(a^{2} - k)}$ $l e t T_{k} = \sqrt{a - \sqrt{k}} \to T_{(a^{2} - k)} = \sqrt{a - \sqrt{a^{2} - k}}$ $∴ I = \underset{k = 1}{\sum^{a^{2} - 1}} \sqrt{a - \sqrt{k}} = \sqrt{2} s_{2}$ $∴ s_{1} + s_{2} = \sqrt{2} s_{2} \to s_{1} = (1 + \sqrt{2}) s_{2} \to \frac{s_{1}}{s_{2}} = (1 + \sqrt{2}) \to ({T h a t}^{'} s i t)$ ${BY Y O R K}$ $T e l e g r a m : b e n g u b l e r$
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