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Questions tagged [geometric-series]

For questions about or involving geometric series, a series where successive terms have a common ratio.

A geometric series is of the form

$$\sum_{n=0}^{\infty}ar^n.$$

If $|r| < 1$, then the series converges to $\frac{a}{1-r}$. If $|r| \geq 1$, the series diverges.

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Geometric Series, Common Difference as relationship between terms

I'm self-studying maths, so have turned to the good people of MSE as I have no teacher to ask. I completed the following problem and didn't find it difficult. However, the answer provided in the text book is different, and I can't see what I've done…
Bangkockney
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How to show $1+r+r^2 + \cdots + r^n = \frac{r^{n+1} - 1}{r-1}$ where $r \neq 1$

I want to show that the geometric series $$1+r+r^2 + \cdots + r^n = \frac{r^{n+1} - 1}{r-1}, r \neq 1$$ I first started with the following using Gauss method: $$S = 1+r+r^2 + \cdots + r^n$$ $$S=r^n + r^{n-1} + r^{n-2} + ... + 1$$ Adding both…
lucidgold
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Are geometric series correctly expressed based on recurrence relation?

He is recurrence relation and my solution: $$ \begin{align} T(n) & = T(n/2) + O(n) \\[6pt] & = T(n/4) + \frac{cn}{2} + cn \\[6pt] & = T(n/8) + \frac{cn}{4}+ \frac{cn}{2} + cn \\ & {}\ \ \ \vdots \\ & = \sum_{i=0}^{logn}\frac{cn}{2^i} = cn…
UserMoon
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converting geometric infinite series to another infinite series

Let $\{x_n\}_{n=1}^\infty$ be a sequence satisfying the recurrence relation: $$ x_n = a\left(1- \sum_{k=0}^{n-1}x_k\right) $$ Where $ x_0 = 1 $, and $a \in [0,1]$ is chosen so that $$ \sum_{k=1}^{\infty} x_k = 1$$ Given a positive integer $d$, how…
SailsMan63
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Is it ok for 'r' to be negative in geometric series?

$6+x+y+z+96 ......$ is a geometric series. Here we need to find the value of $x$. Before doing that we need to find the value of $r$. Here, $a=6$, $ar^4=96$ Now, $r^4=16$, then $r=\pm 2$. But, a group of teacher are saying that here $r$ can't be…
Nur Mohammad Khan Othoy
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Show that $\frac{1-x^n}{1-x}=0\,$ has a real solution at $x$ equals $-1$

I'm trying to solve this question: $$S_n = 1+x+x^2+x^3...+x^{2021}$$ $$xS_n = x+x^2+x^3+x^4...+x^{2022}$$ Subtracting the bottom equation from the top.. $$S_n(1-x)=1-x^{2022}$$ $$S_n=\dfrac{1-x^{2022}}{1-x}$$ Giving that $S_n = 0$, I ended up with…
mjo
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Mathematical Conversion of Geometric Series

Hello I am dealing with a problem where I need to calculate the probability of a player winning a coin toss game where the two players alterantely toss a coin and the first one to toss head wins. It states that the coin does not necessairly need to…
CoffeeKid
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Question regarding geometric series exercice

I have the following sum: $$ S = \sum^{\infty}_{n = 1} \frac{1 + 2^n}{3^n} $$ So I tried the following: $$ S = \sum^{\infty}_{n = 1} \frac{1}{3^n} + \sum^{\infty}_{n = 1} \frac{2^n}{3^n} = \sum^{\infty}_{n = 1} (\frac{1}{3})^n + \sum^{\infty}_{n =…
Norhther
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Calculating weighted geometric returns

Consider the quarterly sales figures and returns in the table below: Qtr Sales ($m) Return (%) Q4 19 8 Q1 20 10 25 Q2 20 13 30 Q3 20 15.6 20 Q4 20 23.4 50 The year on year return for Q4 20 is: $$ Ret = \frac{R_4}{R_1}-1 =…
Pedro Godwand
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Find shortest length of rod divided into 10 pieces in a geometrical progression

A rod one metre in length is divided into $10$ pieces whose lengths are in geometrical progression. The length of the longest piece is eight times the length of the shortest piece. Find, to the nearest millimetre, the length of the shortest…
Steblo
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Find $\mathbb{E}[X(X-1)]$ for the geometric distribution without using derivation

I'm trying to find the $\mathbb{E}[X(X+1)]$ for the geometric distribution. Everywhere I've looked explains how to do it using the derivative but I have not been taught that method. Is there another way to do this? I know that $\mathbb{E}[X(X+1)]…
user229257
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Simplify Geometric series

Can this equation simplify to the property of a sum of a geometric series, such as $ \frac{1}{1-r} $ $$\sum_{y=1}^{\infty}y^2q^{y}p$$ I understand that $$\sum_{y=1}^{\infty}yq^{y} = q \sum_{y=0}^{\infty}(y-1)q^{y-1} = q \frac{d}{dq}…
cappuccino
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Sum of increasing geometric series

Sum of the series $1 + \frac{1+3}{2!}+ \frac{1+3+3^2}{3!}+....... $ The series becomes $ \sum_{k=1}^{\infty} \frac {3^{k-1}}{k!}$. How to calculate it's sum? Is it divergent due to the numerator?
user388189
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The value of $a^{\log_b x}$ where

a=0.2 $b=\sqrt 5$ $x=\frac 14 + \frac 18 + \frac {1}{16}.....$ Since x is a GP, common ratio ‘r’ is $\frac 12$ Then $$x=\frac 12$$ So $$(0.2)^{log_{\sqrt 5}\frac 12}$$ I don’t know how to simply it further. Using a calculator isn’t allowed, and…
Aditya
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Rewriting the nth term of a series in the form $ar^{n-1}$

In the formula below, where does the $\frac{4}{3}$ come from and what happened to the $3$? How did they get the far right answer? Taken from Stewart Early Transcendentals Calculus textbook. $$\sum^\infty_{n=1}…
Jinzu
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