# Determine the sum of the following geometric series. Round your answers to 4 decimals if necessary.a) `sum_{k=1}^50 8(0.5)^{k-2}`

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To sum a geometric series, we need to find the first term, the ratio between terms and the number of terms. Rearranging the series, we see that

`sum_{k=1}^50 8(0.5)^{k-2}`

`=sum_{k=1}^50 8(2)(0.5)^{k-1}`

`=sum_{k=1}^50 16(0.5)^{k-1}`

`=16(1+1/2+1/4+1/8+cdots+1/2^49)`

Now, we see that the ratio between terms is r=1/2, the initial term is 16, and there are 50 terms in the series. The geometric series has the summation formula:

`S={a(1-r^n)}/{1-r}` sub in values

`={16(1-(1/2)^50)}/{1-1/2}` simplify

`=32(1-(1/2)^50)`

If you put this through your calculator, you will get a number that is essentially 32 (to many more than 4 decimal places). However, the actual number is slightly less than 32.

**The sum of the geometric series is 32.**

The value of `sum_(k=1)^50 8*(0.5)^(k - 2)` has to be determined.

`sum_(k=1)^50 8*(0.5)^(k - 2)`= `8*(0.5^-1 + 0.5^0 + 0.5^1 + ... + 0.5^48)`

=> `8*(2 + 1 + 1/2 + 1/4 + ... + 1/48)`

This is a geometric series with the first term 16 and common ratio 1/2. The sum of the first 50 terms is `16*(1 - 0.5^50)/(1 - 0.5) = 32`

**The value of the sum `sum_(k=1)^50 8*(0.5)^(k - 2)` = 32**