Finite geometric and arithmetic series

Adding the integers from 1 through 40, one summand at a time, is a long addition. The row is finite, each neighbor is one more than the last, and the last term is already known. The question is whether that regularity names the sum before the middle terms are touched.

Join each term to its partner

Carl Friedrich Gauss (1777–1855), at nine, was asked to add from 1 through 40. He wrote 820. Write the row twice, the second copy reversed, and add down the columns:

$\begin{array}{ccccc}1& 2& 3& \cdots & 40\\ 40& 39& 38& \cdots & 1\end{array}$

There are 40 columns, and each column adds to 41. The original row is half of that doubled total, $20\cdot 41=820$.

Two extra freezes sit in that story. The first term is 1, and the step from neighbor to neighbor is 1, so the terms are consecutive integers. The board keeps the first term at 1 and shortens the row to eight dots, so the pairs fit on the page. At the start the last dot sits at 8.

Leave $H$ at 8 and the common difference is 1: the dots are 1 through 8. Drag $H$ and those consecutive integers are the special position you just left. What stays is a constant step, eight terms, and every arch still joining two values that add to the first plus the last. Arithmetic series with a constant difference are already in Babylonian and Egyptian texts from about 2000 BCE. The arches are that fact, drawn short.

Half the count times the two ends

Every pair adds to the same number, so the sum is half the number of terms times (first term + last term).

For a first term $a$, a common difference $d$, and a last index $n=1,2,\dots$,

$a+(a+d)+(a+2d)+\cdots +(a+nd)=\frac{n+1}{2}(a+(a+dn)).$

The index $n$ is the last step, so the row has $n+1$ terms, and the last term is $a+nd$. The sum sign writes that row as one symbol,

${\sum }_{k=0}^n{a}_k:=a_0+a_1+\cdots +a_n,$

and the same sum is

${\sum }_{k=0}^n(a+kd)=\frac{n+1}{2}(a+(a+nd)).$

The product sign is the twin abbreviation,

${\prod }_{k=0}^n{a}_k:=a_0{a}_1\cdots a_n.$

The sums on this page never call it.

From 1 through 40, take $a=1$, $d=1$, and $n=39$. Then there are 40 terms, the last term is 40, and

$\frac{40}{2}(1+40)=20\cdot 41=820.$

The doubled list above is those 40 pairs. Half of their total is the series. On the board the same half-count rule still names the sum after $H$ moves; the first term and the number of dots stay frozen. A row whose neighbors do not share one difference has no single pair-sum to halve.

Each bar is the previous bar times $q$

A constant ratio is the other neighbor rule that still closes. The terms are $a$, $aq$, $a{q}^2$, up through $a{q}^n$, with $n=1,2,\dots$, and adjacent terms stand in the ratio $q$. The board draws four bars, so the last power on the figure is 3.

For every real or complex $a$ and $q$ with $q\ne 1$,

$a+aq+a{q}^2+\cdots +a{q}^n=a\frac{1-q^{n+1}}{1-q}.$

With the sum sign,

${\sum }_{k=0}^{n}a{q}^k=a\frac{1-q^{n+1}}{1-q},q\ne 1.$

The bars show the hypothesis this quotient spends: each height is $q$ times the one on its left. Multiply every bar by $q$ and the row steps one place to the right. The two pictures share the middle bars, so the sum is fixed by the first bar and the one new bar at the end. That shift is why the closed form is a difference of powers of $q$, divided by $1-q$. The board drags real $a$ and $q$. The identity is stated for complex values as well.

The short case $a=1$, $n=2$ is three terms,

$1+q+q^2=\frac{1-q^3}{1-q},q\ne 1.$

Euclid's Elements, from about 300 BCE, already records a sum for a geometric series. A finite run of the heights on powers, roots, and logarithms is this series with first term $1$ and ratio equal to the base.

Set the ratio to $1$. Every bar has height $a$, and a row with last index $n$ has $n+1$ of them, so the sum is $(n+1)a$. The quotient above has denominator $1-q$, which is $0$ at $q=1$, so that expression stops there. The equal bars are the sum you can still count.

The object is a finite row. A constant difference gives half the count times the two ends. A constant ratio $q\ne 1$ gives the quotient of powers. In both formulas the index runs from $0$ through $n$, so the number of terms is $n+1$. Further elementary algebraic formulas continue past this page. The question this board can still answer is the equal-bar sum, when $q=1$.