Sum of a geometric series : 1 + r + r^2 + … + r^n = A^2

The general form of a geometric sequence is
$a, \; a\, r, \; a\, r^2, \; a \,r^3, \; a \, r^4, \; ...$

$a$   is the first term, and
$r$   is the factor between the terms (called the “common ratio”)

Find a geometric series of 3 or more positive integers, starting with 1,
such that its sum is a perfect square

that is,

$a \; = \; 1$

$1 \; + \; r \; + \; r^2 \; + \; r^3 \; + \; ... \; + \; r^n \; = \; A^2$

For example,

$1 \; + \; 3 \; + \; 3^2 \; + \; 3^3 \; + \; 3^4 \; = \; 121 \; = \; 11^2$

$1 \; + \; 7 \; + \; 7^2 \; + \; 7^3 \; = \; 400 \; = \; 20^2$

Can you find other examples?

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

2 Responses to Sum of a geometric series : 1 + r + r^2 + … + r^n = A^2

1. paul says:

I can’t find anything else with a^2 <= 331 digit length squares equivalent to r = 2000 and n = 100.

Paul.

• benvitalis says:

I couldn’t find anything else. It’s hard to prove it!