Summation Notation

[nova pond]

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[chrome gazelle]

$a_1 = 1$

[south cave]

do you see the pattern here?

[warm thicket]

xNiden

[chrome gazelle]

$r = \frac{1}{8}$

[warm thicket]

xNiden

[chrome gazelle]

$a_n = a_1r^{n-1}$

[warm thicket]

xNiden

[chrome gazelle]

$\frac{1}{512} = 1\left(\frac{1}{8}\right)^{n-1}$

[warm thicket]

xNiden

[chrome gazelle]

yes the pattern is 1/8 then everything just gets multiplied by 1/8

[south cave]

so the n-th term in general form is?

[south cave]

using this formula, it is $\left(\frac{1}{8}\right)^{n-1}$

[warm thicket]

wolfqz

[chrome gazelle]

oh

bro i made it more complicated than it was

🤦

[onyx saffron]

first term and common ratio?

first like find the first few values

then find a common ratio

[chrome gazelle]

ok

[chrome gazelle]

figured it out

1/8

mb i was doing other hw

$\frac{2^n-3^n}{6^n}$

[warm thicket]

xNiden

[chrome gazelle]

$\frac{2^0-3^0}{6^0}$ = $\frac{1-1}{1}$ = $\frac{0}{1}$

[warm thicket]

xNiden

[chrome gazelle]

$\frac{2^1-3^1}{6^1}$ = $\frac{2-3}{6}$ = $-\frac{1}{6}$

[warm thicket]

xNiden

[chrome gazelle]

$\frac{2^2-3^2}{6^2}$ = $\frac{4-9}{36}$ = $-\frac{5}{36}$

[warm thicket]

xNiden

[chrome gazelle]

$\frac{2^3-3^3}{6^3}$ = $\frac{8-27}{216}$ = $-\frac{19}{216}$

[warm thicket]

xNiden

[chrome gazelle]

$\frac{2^{4}-3^{4}}{6^{4}}$ = $\frac{16-81}{1296}$ $-\frac{65}{1296}$

[warm thicket]

xNiden

[chrome gazelle]

hm

$\frac{1}{1-\frac{2}{6}}$ = $\frac{1}{\frac{4}{6}}$ = $\frac{1}{1}*{\frac{6}{4}}$ = $\frac{6}{4}$

[warm thicket]

xNiden

[chrome gazelle]

$\frac{1}{1-\left(-\frac{3}{6}\right)}$ = $\frac{1}{\frac{3}{6}}$ = $\frac{1}{1}*{\frac{6}{3}}$ = $2$ = $\frac{4}{4}$

$\frac{6}{4}$ + $\frac{4}{4}$ = $\frac{10}{4}$

[warm thicket]

xNiden

[chrome gazelle]

hm

[warm thicket]

xNiden

[chrome gazelle]

$\frac{1}{1-\left(-\frac{3}{6}\right)}$ = $\frac{1}{\frac{9}{6}}$ = $\frac{1}{1}*{\frac{6}{9}}$ = $\frac{2}{3}$

[warm thicket]

xNiden

[chrome gazelle]

im not sure what im doing wrong

[onyx saffron]

i believe theres some way to factor 2^n-3^n

because i dont think you have a good common ratio there

[chrome gazelle]

oh

[chrome gazelle]

@empty dew

[chrome gazelle]

@empty dew

[empty dew]

Yes

Let me see

[onyx saffron]

issue is you used the geometric series fomula but your r value is well

idk something random

[chrome gazelle]

how's it random?

mm

[empty dew]

The answer can't positive for sure

[onyx saffron]

well you can see the ratio clearly doesnt hold

[chrome gazelle]

I tried to figure out the r by setting everything to the power of 0, 1, 2, 3 and etc

im not sure

[empty dew]

That's the peoblwm

[chrome gazelle]

thats true

hm

[onyx saffron]

also your series starts at 3

[empty dew]

It's not geometric

[chrome gazelle]

right

mb

hm

[empty dew]

Oh yeah I didn't see that

[onyx saffron]

believe theres a general way to factor a^n-b^n for odd and even n

[empty dew]

No need for that

[onyx saffron]

whether or not that helps i have no clue

oh ok

[empty dew]

Just add thkse terms when n=0,1,2 and you can aplly it

[empty dew]

I mean to try this

[chrome gazelle]

this is what shows in this yt videohttps://www.youtube.com/watch?v=byY9a0Q1g_g&t=291s

[empty dew]

This way you can combine those 3 extra terms with the sum that starts from 3 into a sum that starts from 0

And the sum on the left I showed you how you can conpute it in the other chat

So after that you just solve for L

[chrome gazelle]

sorry im currently eating brb

[empty dew]

And that's the answee

Bruh

[chrome gazelle]

hm

ok wait ill stay

[empty dew]

No just eat

[chrome gazelle]

so

alright gimme a sec

[empty dew]

Think as much as you want

[chrome gazelle]

like first meal of the day 💀

ill look at it while eating

[empty dew]

Summary: since the sum was starting from 3 we couldn't just mindleslly apply any formulas like the geometric sum formula to help. That's why we forcefully introduced those 3 missing terms to complete the sum and then we could apply those formulas for help

We then break that fraction into 2 fractions and break it into 2 sums which are both geometric

The rest it's just algebra

[chrome gazelle]

So then

hm

i would add $\frac{2^0-3^0}{6^0}$ + $\frac{2^1-3^1}{6^1}$ + $\frac{2^2-3^2}{6^2}$ + $\frac{2^3-3^3}{6^3}$

[warm thicket]

xNiden

[chrome gazelle]

wait you put M= 0 for the Sigma

[chrome gazelle]

hm

im confused

[empty dew]

Why

What does sigma from n = 5 to n = 12 of n^2 ?

It means 5^2 + 6^2 + 7^2 + ... + 12^2

It's just a notation to shorten sums
That's why if we add 1^2 + 2^2 + 3^2 + 4^2 to the sigma from n = 5 to n = 12 of n^2 we can shorten it into just sigma from n = 1 to = 12 of n^2

The same thing I combined the (2^0 - 3^0)/6^0 + ... + (2^2 - 3^2)6^2 with the sum that starts from n = 3 to n = inf and made it into a sum from n = 0 to n = inf

[wicked spindle]

$\sum_{n=3}^{∞} \left( \frac{1}{3^n}-\frac{1}{2^n}\right)$

[warm thicket]

(xy)²=(yz)²=(xz)²=xyyzxz=-1

[wicked spindle]

decompose the fractions

then canceling

$\sum_{n=3}^{∞} \frac{1}{3^n}-\sum_{n=3}^{∞}\frac{1}{2^n}$

[warm thicket]

(xy)²=(yz)²=(xz)²=xyyzxz=-1

[wicked spindle]

then you can take summation inside the additive operator

(add and subtract are both additive operators)

[wicked spindle]

you now have 2 different infinite series

solve each individually

[wicked spindle]

$\frac{2^n - 3^n}{6^n} = \frac{2^n}{6^n} - \frac{3^n}{6^n}
= (\frac{2}{6})^n - (\frac{3}{6})^n$

[warm thicket]

(xy)²=(yz)²=(xz)²=xyyzxz=-1

[wicked spindle]

i thought it was telescopic at first

but when i seperate it

it isn't telescopic but it does help solving the problem

[chrome gazelle]

Sorry Today I got busy

[chrome gazelle]

+close