If this is an infinite sum the formula isNotaPhysicist said:Homework Statement
If
[tex]1 + 2x + 4x^2 + ... = \frac{3}{4}[/tex]
Homework Equations
[tex]S_n = \frac{a(1 - r^n)}{1 - r} [/tex]
[tex] t_n = ar^{n-1} [/tex]
The Attempt at a Solution
a = 1
r = 2x
I try to solve [tex]S_n[/tex] and end up with
[tex]2x^n = \frac{6x - 7}{4}[/tex]
which I can't solve.
I try to solve by equating t2 and t3 and getting x = (1/2). Which is wrong.
Any help appreciated.
HallsofIvy said:The "..." at the end of the sum indicates it is an infinite sum.
A geometric progression is a sequence of numbers where each term is obtained by multiplying the previous term by a constant value. This constant value is called the common ratio and is denoted by the letter "r".
To find the sum of a geometric progression, use the formula S = a(1-r^n)/(1-r), where S is the sum, a is the first term, r is the common ratio, and n is the number of terms. Alternatively, you can use the formula S = a(r^n - 1)/(r-1).
Finding x in a geometric progression is important because it allows us to determine the value of a missing term or the number of terms in a sequence. This information can be useful in various mathematical and scientific contexts.
If the common ratio in a geometric progression is negative, the sequence will alternate between positive and negative values. To find the sum, you can use the formula S = a/(1+r) or S = a(1-r^n)/(1+r), depending on the number of terms.
Yes, a geometric progression can have an infinite sum if the absolute value of the common ratio is less than 1. In this case, the sequence approaches a limit as the number of terms increases, but never reaches it.