Formula for a geometric series with variable common ratio

[ATroelstein]

I have the following summation where a is a positive constant and can be > 1

$$
\sum_{k=2}^{n}a^{n-k}
$$

I am trying to find a general formula for this summation, which turns out to be a geometric series with a as the common ratio. I have worked out the following:

$$
\sum_{k=2}^{n}a^{n-k} = \sum_{k=0}^{n-2}a^{k}
$$$$
a^{0} + a^{1} + a^{2} + ... + a^{n-3} + a^{n-2}
$$Plugging this information into the geometric series summation formula where the common ratio = a, the first term is 1 and the number of terms is n-2, I get:

$$
S_{n-2} = 1 * (\frac{1-a^{n-2}}{1-a})
$$

After testing this formula few times, I found that it's always slightly off. I'm wondering where I went wrong when trying to determine the formula. Thanks.

 

[I like Serena]

Hi ATroelstein! :)

The proper formula is: $S_{n-2} = 1 \cdot \frac {1-a^{n-1}}{1-a}$

See for instance here.

 

[MarkFL]

You want:

$\displaystyle S_{n-2}=\frac{a^{n-1}-1}{a-1}$

Consider:

$\displaystyle S_{n-2}=\sum_{k=0}^{n-2}a^k$

Multiply through by $a$:

$\displaystyle aS_{n-2}=\sum_{k=0}^{n-2}a^{k+1}=\sum_{k=0}^{n-2}a^k-a^0+a^{n-1}=S_{n-2}+a^{n-1}-1$

$\displaystyle aS_{n-2}-S_{n-2}=a^{n-1}-1$

$\displaystyle (a-1)S_{n-2}=a^{n-1}-1$

$\displaystyle S_{n-2}=\frac{a^{n-1}-1}{a-1}$

 

[chisigma]

I have the following summation where a is a positive constant and can be > 1

$$
\sum_{k=2}^{n}a^{n-k}
$$

I am trying to find a general formula for this summation, which turns out to be a geometric series with a as the common ratio. I have worked out the following:

$$
\sum_{k=2}^{n}a^{n-k} = \sum_{k=0}^{n-2}a^{k}
$$$$
a^{0} + a^{1} + a^{2} + ... + a^{n-3} + a^{n-2}
$$Plugging this information into the geometric series summation formula where the common ratio = a, the first term is 1 and the number of terms is n-2, I get:

$$
S_{n-2} = 1 * (\frac{1-a^{n-2}}{1-a})
$$

After testing this formula few times, I found that it's always slightly off. I'm wondering where I went wrong when trying to determine the formula. Thanks.

May be that the best first step is to set... $\displaystyle \sum_{k=2}^{n} a^{n-k} = a^{n}\ \sum_{k=2}^{n} \frac{1}{a^{k}}$ (1)

Kind regards

$\chi$ $\sigma$