5 + 15 + 45 + .... + 98, 415

I think i know how to do this but i dont know the pattern to the numbers and the comma at the end threw me off.:confused:

Any help appreciated. THx.

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- May 9th 2007, 03:36 PMijleytonFind the sums:
5 + 15 + 45 + .... + 98, 415

I think i know how to do this but i dont know the pattern to the numbers and the comma at the end threw me off.:confused:

Any help appreciated. THx. - May 9th 2007, 04:06 PMtopsquark
I think the last number is 98415. Some books still use a comma to separate powers of 3 in the number. (ie. 1000 --> 1,000 and a million is 1,000,000.)

It looks like a geometric series:

a_k = 5*3^{k - 1}

The sum of a geometric series (with the first term equal to 1) is

S_n = (1 - r^{n+1})/(1 - r) = r^0 + r^1 + ... + r^n

where, in this case, r = 3 and n is the label of the last term in the series.

So what is n?

98415 = 5*3^{n - 1}

19683 = 3^{n - 1}

n - 1 = log_3(19683) = ln(19683)/ln(3) = 9

n = 10

So

S_10 = (1 - 3^{10 + 1})/(1 - 3) = 88573

This is the sum for

S_10 = 3^0 + 3^1 + 3^2 + ... + 3^{10}

We need to remember to multiply by 5, so our sum is:

5*88573 = 442865

-Dan