Question

I have this homework and I need to find the formulae for S_n in function of "n" The sum is sum (3)/4^(n+1) n=1 to infinity. I have calculated the first four terms of the series.

These are the first four: 3/16, 3/64,3/256, 3/1024. According my book I have to do:

S1=u1 and S2=S1+u2
S3=S2+u3 S4=S3+u4
As uk=34k+1
Rewriting I got:

uk=34k4 or 3∗4−k−1.

Am I right?? What I can do more from here?

Answer 1

I´m searching for this answer S_n=(4^(n)-1)/4^(n+1).

Answer 2

like this:
\[ \frac{4^n -1}{4^{n+1} } \]
?

Answer 3

And What more?? How can I get to the answer?? Now I know that

S_1=u_1 and S_2=S_1+u_2

S_3=S_2+u3 S_4=S_3+u_4
According to the book u_k=3/4^(k+1)


And by book´s example I have to write:
S_n=u_1+u_2+u_3+u_4+...u_n-1+u_n.

I know that the discretized function that is U_n is (3)/4^(n+1). In this case U_n-1 it will be
(-3)/4^(-n-1)?

Answer 4

And What more?? How can I get to the answer?? Now I know that

S_1=u_1 and S_2=S_1+u_2

S_3=S_2+u3 S_4=S_3+u_4
According to the book u_k=3/4^(k+1)


And by book´s example I have to write:
S_n=u_1+u_2+u_3+u_4+...u_n-1+u_n.

I know that the discretized function that is U_n is (3)/4^(n+1). In this case U_n-1 it will be
(-3)/4^(-n-1)?