# Sum of a geometric series

Someone recently found my blog searching on “1 2 4 8 16 32 64 sequence sum”. They probably found my Pairing Up With Gauss post which didn’t answer their question. If they want a thorough answer, I suggest they google “geometric series” or “geometric sequence” and read pages such as this one that give a good overview of the topic.

If you’d like to read the “middle school approach” to a problem like that one, read on…

For this problem, one should employ the approaches “Solve a Simpler Problem” and/or the related “Find a Pattern”.

Let’s sum the first n terms of that sequence, and keep track of our results.

n |
sequence |
sum |

1 |
1 |
1 |

2 |
1 2 |
3 |

3 |
1 2 4 |
7 |

4 |
1 2 4 8 |
15 |

5 |
1 2 4 8 16 |
31 |

Hopefully you will see the pattern that is emerging. (If not, look closely at each of the sums, and compare it with the next number in the sequence….)

It’s similar but a little more complicated with the powers of three:

n |
sequence |
sum |

1 |
1 |
1 |

2 |
1 3 |
4 |

3 |
1 3 9 |
13 |

4 |
1 3 9 27 |
40 |

5 |
1 3 9 27 81 |
121 |

The pattern is less obvious in this case, but you still need to look at the next number in the sequence. You just have to manipulate it a bit more. (I’ll post the answer in the comments, but try not to give up too quickly!)

Once you get that one, you should be good for figuring out the powers of 4 or higher, or starting from a number other than 1, and coming up with a general formula from there. :)

For the powers of three, the sum of the first n powers of 3 is:

(3^n – 1)/2

(That may look a little different than the usual formula, because I’m talking about “the first n” powers of 3 (which go from 3^0 to 3^(n-1), rather than talking about the sum from 0 to n of 3^n)

I worked out the sum of the first n numbers, and it is:

{n[3^(9n-1)] + (n-2)[3^(n-1) - 3^(n-1) + 2]}/2

I could not simplify as i had quite a bit of difficulty, and that i did not know how to as i am only 13 1/2.