write the following in a sigma notation

3/32+3/16+3/8+...+12288

so far ive gotten that a=3/32 and r=2/1

can you please help me further

sum = [sigma] 3/2^(6-i), i from 1 to 18

what does "i" stand for?

In sigma notation we traditionally use i or n as the index of the notation.

You had asked for the expression in "sigma notation"

What you started doing is attempting to add up the values using a geometric series.
What you started is correct, but that was not asked for.

If you want the actual sum, you will first have to find how many terms there are, that is, what term number is 12288

the use the formula
Sn = a(r^n - 1)/(r-1)

thank you muchly!

Sure! To express the given series in sigma notation, we already know that the first term (a) is 3/32 and the common ratio (r) is 2/1. Now, let's determine the number of terms (n) in the series.

To find the number of terms, we can observe that each term in the series is obtained by multiplying the previous term by the common ratio. Starting with a value of 3/32, we can continue multiplying by the common ratio (2/1) until we reach the final term, which is 12288.

3/32 * (2/1) = 3/16
3/16 * (2/1) = 3/8

Continuing this pattern, we can see that by multiplying each term by 2/1, we eventually reach 12288:

3/8 * (2/1) = 6/8 = 3/4
3/4 * (2/1) = 6/4 = 3/2
3/2 * (2/1) = 6/2 = 3

Therefore, we determined that there are a total of 7 terms in the series.

Now, let's express the series in sigma notation. Since the first term (a) is 3/32, and the common ratio (r) is 2/1, we can use the following formula for the sum of a geometric series:

S = a * (1 - r^n) / (1 - r)

Substituting the known values:

S = (3/32) * (1 - (2/1)^7) / (1 - 2/1)

Simplifying further:

S = (3/32) * (1 - 128/1) / (-1)

S = (3/32) * (-127/1)

Finally, the series can be expressed in sigma notation as:

Σ (3/32) * (-127/1)^k, where k ranges from 0 to 6.