Help Me with this sum serie

[quick bane]

sum:
a + y(a) + y( a + y(a) ) + .... + ?
^
soming c times

the ideia is take all beside terms, then multply it by a "y" number

this pattern repeating "c" times

[quick bane]

<@&286206848099549185>

[royal oasis]

@quick bane

[quick bane]

hi

thanks

[royal oasis]

Are you trying to solve for a specific c value?

[quick bane]

yess

how i can do it?

a = inicial vallue
y = the term keeping multiplying

[royal oasis]

I would encourage you to write out s_1, s_2, and so on

[quick bane]

c = quantity of times

[royal oasis]

Close, as you stated they are sums. So S_2(a) = a + y(a)

[quick bane]

no cause s_2 is the sum of all terms beside times y

so its y * (a)

[royal oasis]

What would you use for s_4?

[quick bane]

y(a + y(a) + y(y(a) + a ) )

[royal oasis]

Okay, I believe I understand better now. I would recommend trying to factor your expressions for these sums. There may be another pattern that can be found.

For example:

S_3 = ay(y+1)

[signal holly]

So, $u_1(a)=a$, $u_n(a)=y\sum_{k=1}^{n} u_k(a)$, and $S_n(a)=\sum_{k=1}^n u_k(n)$

[honest lantern]

SWR

[quick bane]

hm

then Sn(a) = a((1+y)^(n-1)) ?

[signal holly]

how do you get this?

[royal oasis]

Factoring

[quick bane]

a + ya + ya(1+y) + ya(1+y)^2 ... ya(1+y)^(c-1)
a + ya( (1+y) + (1+y)^2 ... + (1+y)^c-1 )
z = 1+y

a + ya( z + z^2 + z^3... z^c-1) = s
(s - a)ya = z + z^2 + z^3... z^c-1
z + z^2 + z^3... z^c-2 = (z^(c-1) - 1 ) /(y)

(s - a)/ya = ((1+y)^(c-1) - 1 )/y
s - a = a( (1 + y)^(c-1) - 1 )
s = a( 1 + (1+y)^(c-1) - 1)
s = a((1+y)^(c-1))

ops

there is c-1

sry

fixed

[signal holly]

Looks good

[quick bane]

other ask

how i create a help

channel?

[signal holly]

this is a help channel

whatever you did to create this

[signal holly]

Or just read #❓how-to-get-help

[quick bane]

i read

[quick bane]

but it doesnt shows

[quick bane]

i mean like this ones

[signal holly]

scroll up a bit more

Or just post in any of these unoccupied ones

[quick bane]

alr

[royal oasis]

I was able to get to the same result that you found. Is that all the question was asking for?

[quick bane]

its not a question

i just thinked

about

this

[royal oasis]

👍

[quick bane]

ill close