/sci/ - Science & Math » Thread #12036012

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Anonymous Sat 22 Aug 15:10:34 2020 No.12036012 View Reply Original Report

Quoted By: >>12036043 >>12036144 >>12037862

QUICK /sci/ !, I need to know as soon as possible what is the formula to calculate these two summations. Please help me, google isn't helping.

a^-i

Anonymous

Anonymous Sat 22 Aug 2020 15:20:27 No.12036043 Report

Quoted By: >>12036060

>>12036012
Rope yourself retard

Anonymous

Anonymous Sat 22 Aug 2020 15:27:12 No.12036060 Report

Quoted By:

>>12036043
thanks /sci/, always helping

Anonymous

Anonymous Sat 22 Aug 2020 15:51:21 No.12036125 Report

Quoted By: >>12036144

I'm not giving you the answer, but I can tell you that both of those approach infinite values if a is greater than one, and 0 if a is less. Figure out your homework/exams yourself retard.

Anonymous

Anonymous Sat 22 Aug 2020 15:59:34 No.12036144 Report

Quoted By:

>>12036012
https://www.mathsisfun.com/algebra/sequences-sums-geometric.html
If you read that page you should be able to figure it out for yourself bud.

>>12036125
That's incorrect, anon

Anonymous

Anonymous Sat 22 Aug 2020 16:51:23 No.12036244 Report

Quoted By:

I am going to assume these sums are multiplied together, because you placed them next to each other
$y = \displaystyle\sum_{i=0}^n{\dfrac{1}{a^i}} \sum_{i=1}^n{\dfrac{1}{a^i}}$
$y = \displaystyle\left(1 + \sum_{i=1}^n{\dfrac{1}{a^i}}\right) \sum_{i=1}^n{\dfrac{1}{a^i}}$
Let $x =\displaystyle\sum_{i=1}^n{\dfrac{1}{a^i}}$
$y = (1 + x) x$
The sum, $x$ , isn't infinite, so it's only possible to get it as a function of both $a$ and $n$ .
$\displaystyle\sum_{i=1}^n{\dfrac{1}{a^i}} = \dfrac{1}{a^1} + \dfrac{1}{a^2} + ... + \dfrac{1}{a^n}$
It's worth noting that this partial sum is the same as the infinite sum minus the infinite sum from n to infinity.
I leave the rest as an exercise for OP

Anonymous

Anonymous Sun 23 Aug 2020 02:49:04 No.12037862 Report

Quoted By:

>>12036012
$<br />\sum^m_{i=m} a_i = a_m \\<br />\sum^n_{i=m} a_i = a_n + \sum^{n-1}_{i=m} a_i<br />$

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