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-Deg

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, \dots

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

\dots

1 + 4 + 8 = 13

the next terms are

4 + 8 + 13 = 25

8 + 13 + 25 = 46

\dots

{it}^{'} s possible to find a zillion more \dots

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 + \dots . . + U_{n}

S_{n} = 1 + 4 + 8 + 13 + \dots . . + U_{n}

S u b t r a c t i n g,

0 = 1 + 3 + 4 + 5 + 6 + \dots . . + (U_{n} - U_{n - 1}) - U_{n}

U_{n} = 1 + 3 + 4 + 5 + 6 + \dots . . + (U_{n} - U_{n - 1})

U_{n} = 1 + 2 + 3 + 4 + 5 + 6 + \dots . . + (U_{n} - U_{n - 1}) - 2

U_{n} = \frac{(n - 1) . n}{2} - 2 \dots \dots (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 .