Geometric Series question

kiwilazer

New member
Joined
Nov 20, 2020
Messages
5
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.
problem#2.PNG
Thanks in advance
 
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
S = n=2133n17n1 \displaystyle S \ = \ \sum_{n=2}^{\infty} \frac{1}{3}*\frac{3^{n-1}}{7^{n-1}} \

S = 17+ n=1173n7n \displaystyle S \ = \ \frac{1}{7} + \ \sum_{n=1}^{\infty} \frac{1}{7}*\frac{3^{n}}{7^{n}} \ ......................................edited

continue.....
 
Last edited by a moderator:
...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
S = n=2133n17n1 \displaystyle S \ = \ \sum_{n=2}^{\infty} \frac{1}{3}*\frac{3^{n-1}}{7^{n-1}} \ will be 1

We can write S = n=0133n+17n+1 \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

S = n=0173n7n \displaystyle S \ = \ \sum_{n=0}^{\infty} \frac{1}{7}*\frac{3^{n}}{7^{n}} \
 
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=17n=2,~a=\frac{1}{7}, the common ratio r=37r=\frac{3}{7} then the sum 17137=14\dfrac{\frac{1}{7}}{1-\frac{3}{7}}=\dfrac{1}{4}
SEE HERE
 
I see this as rather standard geometric series.
First term n=2, a=17n=2,~a=\frac{1}{7}, the common ratio r=37r=\frac{3}{7} then the sum 17137=14\dfrac{\frac{1}{7}}{1-\frac{3}{7}}=\dfrac{1}{4}
SEE HERE
That's what I was going to say!
 
Top