Given the increasing sequence : 1, 4, 8, 13, ... a. Find U_(2019) b. Find S_(2019) U_n is nth−term of the sequence S_n is sum of n − term of the sequence Arithmetic Sequence Degree Two


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Question Number 75818 by naka3546 last updated on 18/Dec/19

$G i v e n t h e i n c r e a s i n g s e q u e n c e :$ $1, 4, 8, 13, . . .$ $a . F i n d U_{2019}$ $b . F i n d S_{2019}$ $U_{n} i s n t h - t e r m o f t h e s e q u e n c e$ $S_{n} i s s u m o f n - t e r m o f t h e s e q u e n c e$ $A r i t h m e t i c S e q u e n c e D e g r e e T w o$
Commented by MJS last updated on 18/Dec/19

$the sequence is not unique$ $1 + 3 = 4$ $4 + 4 = 8$ $8 + 5 = 13$ $the next terms are$ $13 + 6 = 19$ $19 + 7 = 26$ $. . .$ $1 + 4 + 8 = 13$ $the next terms are$ $4 + 8 + 13 = 25$ $8 + 13 + 25 = 46$ $. . .$ ${it}^{'} s possible to find a zillion more . . .$
Commented by MJS last updated on 18/Dec/19

$not really$ $in some cases if we do not know what we are$ $supposed to do we cannot give a proper$ $answer$
Commented by $@ty@m123 last updated on 18/Dec/19

$I t s e x a g g e r a t i o n .$
	Answered by $@ty@m123 last updated on 18/Dec/19
	$S_n =1+4+8+13+.....+U_n S_n = 1+4+8+13+.....+U_n Subtracting, 0=1+3+4+5+6+.....+(U_n −U_(n−1) )−U_n U_n =1+3+4+5+6+.....+(U_n −U_(n−1) ) U_n =1+2+3+4+5+6+.....+(U_n −U_(n−1) )−2 U_n =(((n−1).n)/2) −2 ......(1) U_(2019) =((2018×2019)/2)−2 U_(2019) =1009×2019−2=2037169 S_n =ΣU_n =(1/2)(Σn^2 −Σn)−Σ2 =(1/2){((n(n+1)(2n+1))/6)−((n(n+1))/2)}−2n =((n(n+1))/4){(((2n+1))/3)−1}−2n Put n=2019 & calculate.$
	$S_{n} = 1 + 4 + 8 + 13 + . . . . . + U_{n}$ $S_{n} = 1 + 4 + 8 + 13 + . . . . . + U_{n}$ $S u b t r a c t i n g,$ $0 = 1 + 3 + 4 + 5 + 6 + . . . . . + (U_{n} - U_{n - 1}) - U_{n}$ $U_{n} = 1 + 3 + 4 + 5 + 6 + . . . . . + (U_{n} - U_{n - 1})$ $U_{n} = 1 + 2 + 3 + 4 + 5 + 6 + . . . . . + (U_{n} - U_{n - 1}) - 2$ $U_{n} = \frac{(n - 1) . n}{2} - 2 . . . . . . (1)$ $U_{2019} = \frac{2018 \times 2019}{2} - 2$ $U_{2019} = 1009 \times 2019 - 2 = 2037169$ $S_{n} = Σ U_{n}$ $= \frac{1}{2} (Σ n^{2} - Σ n) - Σ 2$ $= \frac{1}{2} {\frac{n (n + 1) (2 n + 1)}{6} - \frac{n (n + 1)}{2}} - 2 n$ $= \frac{n (n + 1)}{4} {\frac{(2 n + 1)}{3} - 1} - 2 n$ $P u t n = 2019 & c a l c u l a t e .$
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