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A clever solution to find the expected value of a geometric r.v. is those employed in this video lecture of the MITx course "Introduction to Probability: Part 1 - The Fundamentals" (by the way, an extremely enjoyable course) and based on (a) the memoryless property of the geometric r.v. and (b) the total expectation theorem.
I am not sure if there is such thing as testing for convergence a geometric sequence, but I have a problem in my book that asks to test for the convergence or divergence of a sequence: $\lim\limits_{n \to \infty} \frac{2^n}{3^{n-1}}$ Can I use the same method used to test a geometric series for this sequence? So rearranging:
A geometric sequence has its first term equal to $12$ and its fourth term equal to $-96$. How do I find the common ratio? And find the sum of the first $14$ terms
Limit of the geometric sequence. Ask Question Asked 11 years, 9 months ago. Modified 2 years, 11 months ago.
can a geometric sequence have ratio if 1. 0. Is there a specific formula for geometric sequences? 1.
$\begingroup$ The ratio of successive terms is not constant, so this is not a geometric sequence, though it may be related to one $\endgroup$ – J. W. Tanner Commented Mar 17, 2019 at 13:54
Because geometric progressions are based on multiplication, and the most important geometric notion, namely, volume, arises from multiplication (length times width times height). The term “multiplicative” is not used because it already has a special meaning in Number Theory.
I know that in quadratic sequence the second common difference is constant, so I want to apply this idea to geometric sequences, like the second ratio is constant, example : $$ 1, 2, 8 , 64 , 1024 ...
I have a geometric series with the first term 8 and a common ratio of -3. The last term of this sequence is 52488. I need to find the sum till the nth term. While calculating the nth term for 5248...
A geometric sequence $u_n$ is a sequence which is defined recursively by (given $u_1$) $$u_{n+1}=ru_n,\quad r\in \Bbb R $$ where the ratio $r$ is a given number.