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Question Number 163120 by ZiYangLee last updated on 04/Jan/22

	Answered by Rasheed.Sindhi last updated on 04/Jan/22
	$a_1 =7, a_n =((12a_(n−1) )/(37−(a_(n−1) )^2 )) for n≥2 S=a_1 +2a_2 +3a_3 +∙∙∙+2020a_(2020) Sum odigits of S=? ⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣ a_1 =7 a_2 =((12a_1 )/(37−(a_1 )^2 ))=((12∙7)/(37−7^2 ))=−7 a_3 =((12a_2 )/(37−(a_2 )^2 ))=((12(−7))/(37−(−7)^2 ))=7 determinant (((n∈O : a_n =7_( n∈E: a_n =−7^() ) ))) S=a_1 +2a_2 +3a_3 +∙∙∙+2020a_(2020) =7+2(−7)+3(7)+4(−7)+...+2019(7)+2020(−7) =7(1−2+3−4+...+2019−2020) =7{(1+3+5+...+2019)−(2+4+6+...+2020)} =7{((1010)/2)(1+2019)−((1010)/2)(2+2020)} =7∙505(2020−2022)=−7∙505∙2 =−7070 Sum of S 7+0+7+0=14$
	$a_{1} = 7, a_{n} = \frac{12 a_{n - 1}}{37 - {(a_{n - 1})}^{2}} f o r n ⩾ 2$ $S = a_{1} + 2 a_{2} + 3 a_{3} + \cdot \cdot \cdot + 2020 a_{2020}$ $S u m o d i g i t s o f S = ?$ $⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣$ $a_{1} = 7$ $a_{2} = \frac{12 a_{1}}{37 - {(a_{1})}^{2}} = \frac{12 \cdot 7}{37 - 7^{2}} = - 7$ $a_{3} = \frac{12 a_{2}}{37 - {(a_{2})}^{2}} = \frac{12 (- 7)}{37 - {(- 7)}^{2}} = 7$ $\begin{matrix} \underset{\overset{}{n \in E : a_{n} = - 7}}{n \in O : a_{n} = 7} \end{matrix}$ $S = a_{1} + 2 a_{2} + 3 a_{3} + \cdot \cdot \cdot + 2020 a_{2020}$ $= 7 + 2 (- 7) + 3 (7) + 4 (- 7) + . . . + 2019 (7) + 2020 (- 7)$ $= 7 (1 - 2 + 3 - 4 + . . . + 2019 - 2020)$ $= 7 {(1 + 3 + 5 + . . . + 2019) - (2 + 4 + 6 + . . . + 2020)}$ $= 7 {\frac{1010}{2} (1 + 2019) - \frac{1010}{2} (2 + 2020)}$ $= 7 \cdot 505 (2020 - 2022) = - 7 \cdot 505 \cdot 2$ $= - 7070$ $S u m o f S 7 + 0 + 7 + 0 = 14$
	Commented by Tawa11 last updated on 04/Jan/22

	$Great sir .$
	Commented by Rasheed.Sindhi last updated on 04/Jan/22

	$THANX MISS!$
	Commented by mr W last updated on 04/Jan/22

	$n i c e!$
	Commented by Rasheed.Sindhi last updated on 04/Jan/22

	$THANKS SIR!$
	Commented by mr W last updated on 04/Jan/22

	$w h a t i f a_{1} = 1 i n s t e a d o f a_{1} = 7 ?$ $c a n w e f i n d a_{n} = ?$
	Commented by Rasheed.Sindhi last updated on 05/Jan/22

	$a_{1} = 1; a_{n} = \frac{12 a_{n - 1}}{37 - {(a_{n - 1})}^{2}}$ $a_{2} = \frac{12.1}{37 - {(1)}^{2}} = \frac{1}{3}$ $a_{3} = \frac{12 . \frac{1}{3}}{37 - {(\frac{1}{3})}^{2}} = \frac{4}{\frac{333 - 1}{9}} = 4 . \frac{9}{332} = \frac{9}{83}$ $a_{4} = \frac{12 . \frac{9}{83}}{37 - \frac{81}{6889}} = . . . .$ $S i r t h i s i s n o t s o e a s y . I t h i n k w e m u s t$ $f i r s t s o l v e t h e g i v e n r e c u r r e n c e$ $r e l a t i o n a n d o b t a i n a n e x p r e s s i o n$ $f o r a_{n} i n t e r m s o f n . I^{'} l l t r y b u t n o t$ $m u c h c o n f i d e n t!$ $b e g i n : T han k s; g o t o b e g i n$
	Commented by mr W last updated on 05/Jan/22

	$i n d e e d n o t s o e a s y . {l e t}^{'} s t r y i f w e c a n$ $s o l v e f o r a_{n} i n t e r m s o f n .$
	Commented by Rasheed.Sindhi last updated on 05/Jan/22

	$T h e q u e s t i o n {d o e s n}^{'} t b e h a v e s o e a s y$ $f o r a n y v a l u e o t h e r t h a n a_{1} = 7$ $I s t h e r e a n y o t h e r v a l u e o f a_{1} f o r$ $w h i c h a_{n} b e h a v e s p e r i o d i c ?$
	Commented by mr W last updated on 05/Jan/22

	$a_{n} = \frac{12 a_{n - 1}}{37 - {(a_{n - 1})}^{2}} \Rightarrow y = \frac{12 x}{37 - x^{2}}$ $i f y o u w i s h x \to - x, i t m e a n s$ $- x = \frac{12 x}{37 - x^{2}}$ $x^{2} - 37 = 12 \Rightarrow x^{2} = 49 \Rightarrow x = \pm 7$ $i . e . a_{1} = 7 o r - 7 . w e g e t e . g .$ $7, - 7, 7, - 7, . . .$ $i f y o u w i s h x \to x, i t m e a n s$ $x = \frac{12 x}{37 - x^{2}}$ $\Rightarrow x^{2} = 25 \Rightarrow x = \pm 5$ $i . e . a_{1} = 5 o r - 5 . w e g e t e . g .$ $5, 5, 5, 5, . . .$
	Commented by Rasheed.Sindhi last updated on 05/Jan/22

	$S ir, {you}^{'} re deep in maths and hence$ $particularly in this question also!$ $T h a n k s f r o m m y d e e p h e a r t!$
	Commented by mr W last updated on 05/Jan/22

	$t h a n k s!$
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