Geometric Series question

[kiwilazer]

I know how to use the geometric series very well but I am confused because this one problem starts at n=2 which changed the simplified sum. I need some help writing the first term because I simply cant find a good example of this.

Thanks in advance

 

[Deleted member 4993]

I know how to use the geometric series very well but I am confused because this one problem starts at n=2 which changed the simplified sum. I need some help writing the first term because I simply cant find a good example of this.
View attachment 23464
Thanks in advance

\(\displaystyle S \ = \ \sum_{n=2}^{\infty} \frac{1}{3}*\frac{3^{n-1}}{7^{n-1}} \ \)

\(\displaystyle S \ = \ \frac{1}{7} + \ \sum_{n=1}^{\infty} \frac{1}{7}*\frac{3^{n}}{7^{n}} \ \) ......................................edited

continue.....

 

[Steven G]

...OR

You want to start the summation when n=0 and the powers of 3 and 7 to be zero as well, correct?
When you plug in 2 for n the powers of the 3 and 7 in
\(\displaystyle S \ = \ \sum_{n=2}^{\infty} \frac{1}{3}*\frac{3^{n-1}}{7^{n-1}} \ \) will be 1

We can write \(\displaystyle S \ = \ \sum_{n=0}^{\infty} \frac{1}{3}*\frac{3^{n+1}}{7^{n+1}} \ \) since when n=0 the powers will be 1 which is what we need to have.

But when n=0, the power will be 1, not 0. So factor out 3/7 to get

\(\displaystyle S \ = \ \sum_{n=0}^{\infty} \frac{1}{7}*\frac{3^{n}}{7^{n}} \ \)

 

[pka]

I know how to use the geometric series very well but I am confused because this one problem starts at n=2 which changed the simplified sum. I need some help writing the first term because I simply cant find a good example of this.
View attachment 23464

I see this as rather standard geometric series.
First term \(n=2,~a=\frac{1}{7}\), the common ratio \(r=\frac{3}{7}\) then the sum \(\dfrac{\frac{1}{7}}{1-\frac{3}{7}}=\dfrac{1}{4}\)
SEE HERE

 

[Harry_the_cat]

I see this as rather standard geometric series.
First term \(n=2,~a=\frac{1}{7}\), the common ratio \(r=\frac{3}{7}\) then the sum \(\dfrac{\frac{1}{7}}{1-\frac{3}{7}}=\dfrac{1}{4}\)
SEE HERE

That's what I was going to say!

 

[Deleted member 4993]

I was hoping that OP will discover that.