sigma notation [series & sequences]

[keyshawn]

Hi,

I want to double check this math problem for the spring break take-home test:

q: use sigma notation to express the series: 0.2 - .8 + 3.2 - 12.8 + 51.2 - ..........

A: I realized that the powers of 2 are at odd integers, then divided by 10.

[i.e. 2^1=2
2^3=8
etc.....]

but I'm trying to figure out how to express this [the fact only odd integers are used as the powers] in notation....

so far i have:

   [infinity symbol]          [2^k] (-1)^(k+1)
 [Sigma symbol]                       _____________________
  K= 1                                                  10

thanks for the help,
will.

[elsesomebody]

Well my first suggestion would be to put the series {1,3,5,7,9...} in sigma notation. These are the exponents you're raising two to. This is what you're hung up on, and hopefully by breaking it down this way you can do it.

[iccky]

Originally Posted by keyshawn

so far i have:

   [infinity symbol]          [2^k] (-1)^(k+1)
 [Sigma symbol]                       _____________________
  K= 1                                                  10

Almost there, but like you say the formula you have now will give you .2-.4+.8 ...

Try this:

   [infinity symbol]          [2^(2k+1)] (-1)^(k)
 [Sigma symbol]                       _____________________
  K= 0                                                  10

I'm a bit rusty and working quickly but that should work.

[elsesomebody]

Originally Posted by iccky

Almost there, but like you say the formula you have now will give you .2-.4+.8 ...

Try this:

   [infinity symbol]          [2^(2k+1)] (-1)^(k)
 [Sigma symbol]                    _____________________
  K= 0                                  10

I'm a bit rusty and working quickly but that should work.

Actually I believe it should be:

   [infinity symbol]          [2^(2k-1)] (-1)^(k+1)
 [Sigma symbol]                _____________________
  K= 0                               10

[keyshawn]

aha - that's it !

thanks for your help ickky and elsesomebody !


catcha back on the flipside,
keyshawn

[Hain]

Well first of this cant be an infinite series since the sum is divergent. And also the equations presented before are somewhat difficult to work with when dealing with a sum.

 n
Sum   -((-4)^k)/20
k=1

Then the sum just becomes

remember: a1*(1-r^n)/(1-r) ; where r is -4 and a1 is the first term .2
          so:
(1-Tn)/25 ; Where Tn is the finishing term of your sum

[iccky]

Originally Posted by keyshawn

aha - that's it !

thanks for your help ickky and elsesomebody !


catcha back on the flipside,
keyshawn

No problem

I always wanted to answer a math problem in tilted knowledge but all these smart TFPers always beat me to it.