Question

How do I write the series in summation notation for 1-3+9-27+81......

Answer 1

Do you know the pattern?

Answer 2

hint: \[\sum_{i=1}^{n} a^i......\]

Answer 3

\[a _{i}\] actually.
To find \[a _{i}\] identify a pattern.

Answer 4

no its a^i

Answer 5

the index here is an exponent

Answer 6

It's geometric I believe ?

Answer 7

yes and it takes advantage of the fact that a negative times a negative equals a positive

Answer 8

Yes , so now I'm stuck on how to solve it ... It's confusing at least right now it is

Answer 9

well its easy to see that we are involved with powers of 3 here...correct?

Answer 10

\[\sum_{i = 1}^{\infty}\]
This is what we know so far

Answer 11

Yes ....

Answer 12

We are trying to find the formula, to do this we must identify a pattern

Answer 13

\[\sum_{i=1}^{n} -3^i\]

Answer 14

check that and make sure you completely understand why it is correct

Answer 15

Pattern as in r= -3 right /; ?

Answer 16

Oh so I am correct ?

Answer 17

Yes, -3 but you multiply.

Answer 18

To what exactly ?

Answer 19

you need to read more about summation (sigma) notation. heres a good site http://tutorial.math.lamar.edu/Classes/CalcI/SummationNotation.aspx

Answer 20

If it was \[-3^{i}\] then then \[a_{2}\] would be -9 and this is not correct.

Answer 21

no because -3*-3 =9

Answer 22

\[\sum_{i=1}^{n} (-3)^i\]

Answer 23

Yes it works with ()

Answer 24

Sorry you are right it is exponential.

Answer 25

Ok thanks & for the site too , @Edutopia & @af0998

Answer 26

np here is another site that explains things simpler http://www.themathpage.com/aprecalc/sigma.htm

Answer 27

Sum is S=1-3+9-27+.....
or S=(1+9+81+....)-(3+27+243+....)
or S=(a geometric series with a=1,r=9 upto n terms)-(a geometric series with a=3,r=9 upto n terms).
or \[S=\sum_{n=0}^{n} 3^{2n}-\sum_{n=0}^{n}3^{2n+1}\]

Another method:
as suggested by @Edutopia