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Can anyone help solve this geometric series? 3 + 6 + 12 + ... 768

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brandih | eNotes Employee

Posted September 3, 2013 at 11:26 PM via web

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Can anyone help solve this geometric series?

3 + 6 + 12 + ... 768

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steveschoen | College Teacher | (Level 3) Assistant Educator

Posted September 3, 2013 at 11:47 PM (Answer #1)

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Hi, Brandi,

For this, we need the formula:

Sn = a1(1-r^n)/(1-r)

Sn = the sum, what we are looking for

a1 = first term

r = common ratio between terms

n = how many terms there are

We have everything but n.  But, we can calculate that with:

an (read a sub n) = a1 * r^(n-1)

For us, an would be the last term, since we are looking for the total number of terms.  The number of terms is n.

We have an and a1.  We know the common ratio is 2 (6/3 = 2, 12/6 = 2).  So, subbing these numbers in:

768 = 3 * 2^(n-1)

Divide each side by 3:

256 = 2^(n-1)

The left side can be written as:

2^8 = 2^(n-1)

So, we would have:

8 = n-1

So, n = 9, meaning there would be 9 total terms.

Going back to the first part:

Sn = a1(1-r^n)/(1-r)

We now have a1, r and n.  So, subbing these numbers in:

Sn = 3(1 - 2^9)/(1-2) = 3(1-512)/(-1) = 3*(-511)/(-1) = 1533


That's the answer.  The sum = 1533

Good luck, Brandi.  I hope this helps.

Till Then,

Steve

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