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Question Number 213802 by efronzo1 last updated on 17/Nov/24

	Answered by A5T last updated on 17/Nov/24

	$t_{1} = k - 1; t_{2} = k$ $t_{3} = \frac{5 k + 1}{25 k - 25}; t_{4} = \frac{\frac{10 k - 4}{5 k - 5}}{25 k} = \frac{10 k - 4}{125 k (k - 1)}$ $t_{5} = \frac{\frac{- 15 k - 4 + 25 k^{2}}{25 k (k - 1)}}{\frac{5 k + 1}{k - 1}} = \frac{25 k^{2} - 15 k - 4}{25 k (5 k + 1)} = \frac{(5 k + 1) (5 k - 4)}{25 k (5 k + 1)}$ $\Rightarrow t_{5} = \frac{5 k - 4}{25 k}; t_{6} = \frac{\frac{10 k - 4}{5 k}}{\frac{10 k - 4}{5 k (k - 1)}} = k - 1 \Rightarrow t_{7} = \frac{5 k - 4}{\frac{5 k - 4}{k}} = k$ $\Rightarrow (t_{1}, t_{2}, . . ., t_{5}) = (t_{6}, t_{7}, . . ., t_{10}) = . . . = (t_{5 n - 4}, t_{5 n - 3}, . . ., t_{5 n})$ $2020 = 5 n \Rightarrow t_{2020} = t_{5} = \frac{5 (21) - 4}{25 \times 21} = \frac{101}{525}$
	Answered by Frix last updated on 17/Nov/24

	$n ⩾ 1$ $t_{5 n - 4} = a$ $t_{5 n - 3} = b$ $t_{5 n - 2} = \frac{5 b + 1}{25 a}$ $t_{5 n - 1} = \frac{5 a + 5 b + 1}{125 a b}$ $t_{5 n} = \frac{5 a + 1}{25 b}$
	Answered by golsendro last updated on 17/Nov/24
	$t_n = ((5t_(n−1) +1)/(25t_(n−2) )) ⇒5t_n = ((5t_(n−1) +1)/(5t_(n−2) )) set a_n = 5t_n ⇒ { ((a_1 =5×20=100)),((a_2 =5×21=105)) :} a_n = ((a_(n−1) +1)/a_(n−2) ) ⇒a_3 = ((53)/(50)) ; a_4 = ((103)/(5250)) a_5 = ((101)/(105)) ; a_6 = 100 , a_7 = 105 a_n o$
	$t_{n} = \frac{{5 t}_{n - 1} + 1}{{25 t}_{n - 2}} \Rightarrow {5 t}_{n} = \frac{{5 t}_{n - 1} + 1}{{5 t}_{n - 2}}$ $set a_{n} = {5 t}_{n} \Rightarrow {\begin{cases} a_{1} = 5 \times 20 = 100 \\ a_{2} = 5 \times 21 = 105 \end{cases}$ $a_{n} = \frac{a_{n - 1} + 1}{a_{n - 2}} \Rightarrow a_{3} = \frac{53}{50}; a_{4} = \frac{103}{5250}$ $a_{5} = \frac{101}{105}; a_{6} = 100, a_{7} = 105$ $a_{n} \underset{―}{\underset{⏟}{o}}$
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