A geometric series represents the partial sums of a geometric sequence. The nth term in a geometric series with first term a and common ratio r is:T(n) = a(1 - r^n)/(1 - r)
No.
Divide any term in the sequence by the previous term. That is the common ratio of a geometric series. If the series is defined in the form of a recurrence relationship, it is even simpler. For a geometric series with common ratio r, the recurrence relation is Un+1 = r*Un for n = 1, 2, 3, ...
a sequence of shifted geometric numbers
A descending geometric sequence is a sequence in which the ratio between successive terms is a positive constant which is less than 1.
antonette taño invented geometric sequence since 1990's
A geometric series represents the partial sums of a geometric sequence. The nth term in a geometric series with first term a and common ratio r is:T(n) = a(1 - r^n)/(1 - r)
The geometric series is, itself, a sum of a geometric progression. The sum of an infinite geometric sequence exists if the common ratio has an absolute value which is less than 1, and not if it is 1 or greater.
No.
Yes, that's what a geometric sequence is about.
A geometric series.
Divide any term in the sequence by the previous term. That is the common ratio of a geometric series. If the series is defined in the form of a recurrence relationship, it is even simpler. For a geometric series with common ratio r, the recurrence relation is Un+1 = r*Un for n = 1, 2, 3, ...
This is not a geometric series since -18/54 is not the same as -36/12
An arithmetic series is the sequence of partial sums of an arithmetic sequence. That is, if A = {a, a+d, a+2d, ..., a+(n-1)d, ... } then the terms of the arithmetic series, S(n), are the sums of the first n terms and S(n) = n/2*[2a + (n-1)d]. Arithmetic series can never converge.A geometric series is the sequence of partial sums of a geometric sequence. That is, if G = {a, ar, ar^2, ..., ar^(n-1), ... } then the terms of the geometric series, T(n), are the sums of the first n terms and T(n) = a*(1 - r^n)/(1 - r). If |r| < 1 then T(n) tends to 1/(1 - r) as n tends to infinity.
a sequence of shifted geometric numbers
A geometric sequence is : a•r^n while a quadratic sequence is a• n^2 + b•n + c So the answer is no, unless we are talking about an infinite sequence of zeros which strictly speaking is both a geometric and a quadratic sequence.
A descending geometric sequence is a sequence in which the ratio between successive terms is a positive constant which is less than 1.