Thread: Write the series using summation notation

1. Write the series using summation notation

3. a) Write the series 60 -15 +15/4 -15/16 +15/64 -15/256.

My work:

I discovered each term is divided by -4.

60 / -4 = -15
15 / -4 = 15/4
15/4 / -4 = -15/16

Total terms 6 (above sigma, also n)
First term k=1 (first term)

Now for the part on the right of sigma, I have difficulty with this.

I wrote the part on the right of sigma, which should be {a+(n-1)d}, as {60+95) / -4}.

I've never encountered a series where everything is divided rather than multiplied so I'm a little lost.

2. Originally Posted by thekrown
3. a) Write the series 60 -15 +15/4 -15/16 +15/64 -15/256.

My work:

I discovered each term is divided by -4.

60 / -4 = -15
15 / -4 = 15/4
15/4 / -4 = -15/16

Total terms 6 (above sigma, also n)
First term k=1 (first term)

Now for the part on the right of sigma, I have difficulty with this.

I wrote the part on the right of sigma, which should be {a+(n-1)d}, as {60+95) / -4}.

I've never encountered a series where everything is divided rather than multiplied so I'm a little lost.

This is geometric...

$\sum_{n=0}^{\infty}60\left(\frac{-1}{4}\right)^n$

3. My question only lists 6 terms for the series. Would it be okay to list n=1 and in the right side of sigma have n-1 instead of just n?

Second, with a 6 term series would the infinity sign then be 6?

4. Originally Posted by thekrown
My question only lists 6 terms for the series. Would it be okay to list n=1 and in the right side of sigma have n-1 instead of just n?

Second, with a 6 term series would the infinity sign then be 6?
Sorry...

Correction

$\sum_{n=1}^660\left(\frac{-1}{4}\right)^{n-1}$

5. Thank you. Really helped a lot. Cheers!

6. I just reviewed this question and get -4 as the 'r' value not -1/4. Can you help clarify this issue?