Question

How do you find the sum of the first six terms of the geometric sequence 1, 1.5, 2, 2.25, ...?

Answer 1

This is not a geometric sequence. The term `2` should not be there.

Without the `2`, the first `6` terms of the geometric sequence sum to `20.78125`

Explanation:

Discarding the errant term `2`, the sequence `1`, `1.5`, `2.25` is a geometric sequence with common ratio `1.5`.

The general formula for a term of a geometric sequence is:

`a_n = a r^(n-1)`

where `a` is the initial term and `r` is the common ratio.

For our sequence `a=1` and `r = 1.5`

Then:

`(r-1) sum_(n=1)^N a_n`

`=(r-1) sum_(n=1)^N a r^(n-1)`

`=r sum_(n=1)^N a r^(n-1) - sum_(n=1)^N a r^(n-1)`

`=sum_(n=2)^(N+1) a r^(n-1) - sum_(n=1)^N a r^(n-1)`

`= a r^N + color(red)(cancel(color(black)(sum_(n=2)^N a r^(n-1)))) - a - color(red)(cancel(color(black)(sum_(n=2)^N ar^(n-1))))`

`= a(r^N-1)`

So:

`sum_(n=1)^N a_n = (a(r^N-1))/(r-1)`

So for our sequence:

`sum_(n=1)^6 a_n = (1(1.5^6-1))/(1.5-1)`

`=2((3/2)^6-1)`

`=2(729/64-1)`

`=(729-64)/32`

`=665/32`

`=20.78125`