Is there a name for the sum of increasing powers?

Question

I was thinking about the wheat and chessboard story and thinking of the total number of grains of wheat…

$$\sum_{n=0}^{63} 2^n$$

And wondered if there is a name for a sum like this?

$$2^0 + 2^1 +2^2 + 2^3 + \cdots$$

Answer 1

This is called a geometric sum/series according as $N$ is finite or not. More generally, it is of the form $$ \sum_{n=0}^Nar^n$$ where $N$ may be "equal to" infinity. In the case where $N$ is infinite, if the series is to converge, we require $\lvert r\rvert<1$. Furthermore, for $N$ finite $$ \sum_{n=0}^Nar^n=a\times\frac{1-r^{N+1}}{1-r},\:\:\:\: r\ne1$$ and $$ \sum_{n=0}^\infty ar^n=\frac{a}{1-r},\:\:\:\:\lvert r\vert<1.$$

Answer 2

It's mainly a well known value $2^{n+1}-1$ . Write it in base $2$.

It is also OEIS sequence A000225 :

$2^n - 1.$ (Sometimes called Mersenne numbers, although that name is usually reserved for A001348.)

Answer 3

As you are looking for novel names, this is also a simple case of the more generic notion of hypergeometric series $$\sum_{k}r_k \,,$$ with $r_0 = 1$, and the ratio of two consecutive terms is a rational function, a ratio of two polynomials $P$ and $Q$ in the summation index $k$

$$ \frac{r_{k+1}}{r_k}= \frac{P(k)}{Q(k)}\,. $$

In your case, you can choose $P$ and $Q$ such that their ratio is equal to $2$. When the ratio is constant, it is called a geometric series (as answered here). As a reminder, it is a sum of terms in geometric progression (se.math) like $1,r,r^2,r^3,\ldots$, whose name (the geometry part) is illustrated by the following figure:

Hypergeometric series are also connected to chess. A rook is a move on a chessboard. Some have developed studies some types of permutations as the placement of a number of rooks on a chessboard-like grid, see for instance Rook theory and hypergeometric series, J. Haglund, 1996.

Answer 4

The formula for the sum of the first $N+1$ terms can be derived as following:

Let:

$$S(N)=\sum_{n=0}^{N} r^n$$

Then distributing an $r$ (which is independent of $n$) along the sum we have,

$$rS(N)=\sum_{n=0}^{N} r^{n+1}$$

Note (1):

$$S(N+1)=\sum_{n=0}^{N+1} r^n=r^0+\sum_{n=0}^{N} r^{n+1}$$ $$=1+rS(N)$$

Also note,

$$S(N+1)-S(N)=\sum_{n=0}^{N+1} r^n-\sum_{n=0}^{N} r^n=r^{N+1}$$

As all terms but $r^{N+1}$ will reduce each other out to zero.

So,

$$S(N+1)=S(N)+r^{N+1}$$

Hence we have from (1) and from the above:

$$S(N)+r^{N+1}=rS(N)+1$$

$$r^{N+1}-1=rS(N)-S(N)$$

$$r^{N+1}-1=(r-1)S(N)$$

Thus,

$$S(N)=\frac{r^{N+1}-1}{r-1}$$

Now we multiply by a special form of $1$ to get:

$$S(N)=\sum_{n=0}^{N} r^n=\left( \frac{-1}{-1} \right) \left( \frac{r^{N+1}-1}{r-1} \right)=\frac{1-r^{N+1}}{1-r}$$

And hence (distributing out an $a$ which is independent of $n$) we get the formula for the geometric sum,

$$\sum_{n=0}^{N} ar^n=a\frac{1-r^{N+1}}{1-r}$$

Note though our formula only works for $r \neq 1$ (our manipulations are okay until we get to dividing by $0=1-r=1-1$ for $r=1$).