MHB Understanding Complex Geometric Sequences: A Revision Question

AI Thread Summary
The discussion focuses on solving a revision question related to complex geometric sequences. The sequence is defined as \( u_n = 3(1+i)^{n-1} \), leading to the product sequence \( v_n = u_n u_{n+k} = 9(1+i)^{2n+k-2} \). It is established that \( v_n \) is also a geometric sequence, as shown by the consistent ratio \( \frac{v_n}{v_{n-1}} = (1+i)^2 \). The conversation emphasizes the importance of understanding the formulas for the sum of geometric sequences for further questions. Additional resources, such as Wikipedia, are suggested for further clarification.
paul6865
Messages
1
Reaction score
0
need a hand with a revision question, I don't quite understand how to go about solving it question is attached below
View attachment 5916
 

Attachments

  • Screen Shot 2016-08-25 at 4.55.45 PM.png
    Screen Shot 2016-08-25 at 4.55.45 PM.png
    23 KB · Views: 114
Mathematics news on Phys.org
Hi, and welcome to the forum!

According to the definition of $u_n$,
\[
u_4=(1+i)u_3=(1+i)^2u_2=(1+i)^3u_1=3(1+i)^3=-6+6i.
\]
In general, $u_n=3(1+i)^{n-1}$. From there
\[
v_n=u_nu_{n+k}=3(1+i)^{n-1}\cdot3(1+i)^{n+k-1}=9(1+i)^{2n+k-2}.
\]
In particular,
\[
v_{n+1}=9(1+i)^{2n+k}=(1+i)^2v_n
\]
which means that $v_n$ is also a geometric sequence.

Another way to show this is to note that $\dfrac{u_n}{u_{n-1}}=1+i$ for all $i$. Therefore
\[
\dfrac{v_n}{v_{n-1}}=\dfrac{u_nu_{n+k}}{u_{n-1}u_{n+k-1}}=
\dfrac{u_n}{u_{n-1}}\cdot\dfrac{u_{n+k}}{u_{n+k-1}}=(1+i)(1+i)=(1+i)^2.
\]
The ratio does not depend on $n$; therefore, $v_n$ is a geometric sequence.

To answer other questions, you need to know formulas for the sum of geometric sequences and other information, which you can find in Wikipedia.

Please also read the http://mathhelpboards.com/rules/, especially rules 8 and 11.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Replies
3
Views
2K
Replies
2
Views
2K
Replies
2
Views
2K
3
Replies
108
Views
10K
Replies
7
Views
2K
Back
Top