Question

Can someone explain how to use sigma notation to write a geometric series? Pleasee

Answer 1

\[\sum_{i=0}^na r^i\]

Answer 2

does that work for every series?

Answer 3

every series that starts from i=0, yes

Answer 4

n could be infinity, and you could start at a different n, but it would still be a geometric series

Answer 5

what if it's like i= 1?

Answer 6

that is an index shift, if you don't want to change the value of the series that means you must subtract 1 from i in the summand, so\[\sum_{i=0}^na r^i=\sum_{i=1}^nar^{i-1}\]

Answer 7

in general\[\sum_{i=p}^na r^i=\sum_{i=p+q}^nar^{i-q}\] that is how index shifts are done

Answer 8

what does the q and p stand for :/ sorry

Answer 9

p is just any starting value for i
for the next part, if you want to start at some different value for i, say p+q instead of p (both p and q are unknown, but that doesn't matter) you have to subtract q from each i in the summand. notice this works if q is negative too (i.e we want to start from an earlier value of i) that means we will be adding q to each i that appears in the summand.

Answer 10

why are we adding q though ?

Answer 11

take my earlier example:\[\sum_{i=0}^na r^i=\sum_{i=1}^nar^{i-1}\]why did we need to add 1 if we change the initial value of i from 1 to 2?
try to find the first 3 terms of each sequence and you should see why this works

Answer 12

*why did we need to subtract 1 if we change the initial value of i from 0 to 1?
try to find the first 3 terms of each sequence and you should see why this works

Answer 13

first term of\[\sum_{i=0}^nar^i\]is\[ar^0=a\]the first term of\[\sum_{i=1}^nar^{i-1}\]is\[ar^{1-1}=ar^0=a\]so by changing the starting index value of i we need to either add or subtract from the i's in the summand to make the series come out the same

Answer 14

to generalize and get what I wrote earlier, let 0=p and 1=q
then you get\[\sum_{i=p}^na r^i=\sum_{i=p+q}^nar^{i-q}\]

Answer 15

I get it, thank you so much you're a life saver !

Answer 16

anytime, happy to help :)