Geometric sequences

Elementary combinatorics: Arithmetic and geometric sequences

Geometric sequences

Geometric row

A geometric sequence with ratio $\boldsymbol{r}$ is a sequence of numbers $a_0,a_1,a_2, \ldots$ for which the quotient $a_{k+1}/a_k$ of two consecutive terms of the sequence is equal to the constant $r$ . So each term in the sequence, except the first one, is the result of multiplication of its predecessor by $r$ : $a_{k+1}=r\cdot a_k$ .

Examples

$\begin{aligned}&1,2,4,8,16,\ldots\\[0.25cm] &2,6,18,54,162,\ldots\end{aligned}$

Sum formula for a geometric sequence

If $a_0,a_1,a_2\ldots$ is a geometric sequence with ratio $r$ , then the partial sum of the first $n$ terms is given by

$\begin{aligned}\sum_{k=0}^{n-1}a_k&= a_0\cdot \sum_{k=0}^{n-1}r^k\\[0.25cm] &=a_0\cdot \frac{1-r^n}{1-r}\end{aligned}$

Examples

$\begin{array}{rcrcrcrcrrc} 1&+&2&+&4&+&8&+&16&&\\ &&2&+&4&+&8&+&16&+&32\\ \hline \end{array}$ So: $\displaystyle \sum_{k=0}^{4}2^k=\frac{1-32}{1-2}=2^5-1=31$

If $a_1=3, r=2, n=20$ , then $\phantom{\qquad\qquad\qquad}$

$\begin{aligned}\sum_{k=0}^{19}(3\times 2^k) &=3\times\frac{1-2^{20}}{1-2}\\[0.25cm]&=3\times (2^{20}-1)\\[0.25cm] &=1048575\end{aligned}$

The proof of the sum formula for the geometric sequence $a_0$ , $a_0 r$ , $a_0 r^2$ , $\ldots$ , i.e. for $a_k=a_0 r^k$ , goes as follows. Denote $\displaystyle s_n=\sum_{k=0}^{n-1} a_0 r^k$ . Then: $r\, s_n=\sum_{k=1}^{n} a_0 r^k$ and therefore $(1-r)s_n=a_0-a_0r^n)$ or $s_n=a_0(1-r^n)/(1-r)$ .

Starting with row index 1 You can also start the indexing of a sequence of numbers with $1$ . The sum formula then becomes

$\begin{aligned}\sum_{k=1}^{n}a_k&=\sum_{k=1}^{n}a_1r^{k-1}\\[0.25cm] &=a_1\frac{1-r^{n}}{1-r}\end{aligned}$ with $r$ the quotient between two consecutive terms from the sequence. You can see that the formula is pretty much the same as the one with indexing from $0$ . But it is a bit more difficult to describe the individual terms mathematically and the indexing does not fit the exponent of the ratio very well.