Chapter 22. Geometric and
Arithmetic Sums and Sequences
Volume 2, Three Skills for Algebra
Addition Formulas
This chapter is required for the comprehension of the next chapter
Personal Money Computations.
1 Sequences
The list
1, 3, 5, 7, 8, 10, 20, 555, 2, 1
forms a 10 member sequence of numbers. More generally, a list of
numbers $f(1), f_2, f_3, f_4, \ldots $ is called a sequence. This sequence
may be given explicitly by a list of numbers. In this case, $f(1)$ is
the first element in the list, $f(2)$ is the second element
in the list, and $f(j)$ is the $j$-th element - provided the
list or sequence is long enough.
The $j$-th term $f(j)$ in a sequence is often
written as $a_j$. Here $a_j$ is given by some formula or
function. Thus we may write
$a_j=f(j)$ or $a_1,a_2,a_3,\dots$ as the shorthand
representation for a sequence. Also, we may employ $b_j$
or $c_j$ etc in place of $a_j$.
2 Arithmetic Sequences and Sums
An example of an eight-member sequence is given by 3, 3.5, 4, 4.5, 5,
5.5, 6, 6.5, and 7. The $j$-th
term in this sequence is given by $a_j=3+j\cdot\frac12$ --
verify this for $j=$ 0, 1, 2, \ldots, 8. The
difference between adjacent or successive members $d=a_{j+1}-a_j$ is the constant
$\frac12$.
According to the next definition, this eight member
sequence is an arithmetic sequence.
Definition: A sequence of numbers
$a_1,a_2,a_3,a_4,\ldots$ is an arithmetic sequence if and
only if there is a constant $d$ such that adding $d$ to
each term in the sequence yields the next term in the
sequence. More briefly, $a_{j+1}=a_j+d$ for some constant
$d$ which does not depend on $j$.
2.1 Arithmetic Sums
First Example: Suppose we want to compute the following sum
\[S=11+14+17+20+23+26+29\] of the seven numbers: $a_1=11$,
$a_2=14=a_2+3$, $a_3=17=a_2+3$, $a_4=20=a_3+3$, $a_5=23=a_4+3$,
$a_6=26=a_5+3$ and $a_7=29=a_6+3$. Except for the first number in this
sequence, each number $a_k$ is given by adding the constant $3$ to its
predecessor $a_{k-1}$, that is, the number immediately before it. To find
the sum $S$, write it forward and backwards as follows. \begin{eqnarray*}
& &S=11+14+17+20+23+26+29 \\
&{+}&\underline{S=29+26+23+20+17+14+11} \\
&&\llap{2}{S=40+40+40+40+40+40+40} \end{eqnarray*} The alignment
of the forward and backward written sums in seven columns, count them,
suggests $2S=7(40)$. This implies \[S=\frac12 \cdot 7(40)=140\] Observe
that $40=29+11$ is the sum of the first and last terms in the sequence.
Also observe that the terms of the sum written forward are increasing by
3 while the terms written backwards are decreasing by 3. This indicates
why the seven columns all give the same number, here 40, when added.
A Second Example: Suppose we want to compute the following sum
Suppose we want to compute the following sum
\[S=11+12+13+14+15+16+17+18+19\] of nine numbers $b_1 =11 $, $b_2
=12=b_1+1 $, $b_3 =13=b_2+1 $, $b_4 =14=b_3+1 $, $b_5 =15=b_4+1 $, $b_6
=16=b_5+1 $, $b_7 =17=b_6+1 $, $b_8 =18=b_7+1 $ and $b_9 =19=b_8+1 $.
Except for the first number, each number is obtained by adding 1 to the
number before it. A quick way to obtain the sum is to write the sum
forward and backwards: \begin{eqnarray*} &&S=11 +12 +13 +14 +15
+16 +17 +18 +19 \\ &{+}&\underline{S=19 +18 +17 +16 +15 +14 +13
+12 +11}\\ && \llap{2}{S=30 +30 +30 +30 +30 +30 +30+30 +30} \\
\end{eqnarray*} Since there are nine terms in the sum, there are nine
columns. Therefore $2S=9\times 30$. \[S=\frac12 \cdot 9(30)=\frac12\cdot
270=135\] Here $30=11+19$ is the again the sum of the first and last
terms. It also the sum of the second and second to last terms and so on.
That is the column sums are again constant. They are constant because the
sequence of terms in the sum form an arithmetic. The above arithmetic sum
examples follow the pattern: \[ 2S=\mbox{(Number of Terms)}
\mbox{(First Term + Last Term)}\] From this pattern, we conclude the sum \[ S=
\frac12\cdot\mbox{(Number of Terms)}\mbox{(First Term+Last Term)}\]
Exercises. Explain why
- the sum of the first 500 whole numbers 1+2+3 + ¼ + 488 +499+500 equals 250·501.
- the sum S = 2+4+5+6+8 is not given by \[ S=
\frac12\cdot\mbox{(Number of Terms)}\mbox{(First Term+Last Term)}\]
3 Geometric Sequences and Sums
A sequence of numbers $c_1,c_2,c_3,c_4,\ldots$ is called an geometric
sequence if there is a constant $a$ such that multiplying each term in the
sequence by $a$ yields the next term in the sequence. That is,
$c_{j+1}=c_j\cdot a$ for some constant multiplier $a$ which does not depend
on $j$. An example of an arithmetic sequence is given by 6, 12, 24, 48, 96,
186 and so on. The $j$-th term in this sequence is given by $c_j=3\cdot
2^j$. The constant multiplier here is $a=2$. Three more examples follow:
\begin{eqnarray*} 1,2,4,8,16,32,64,128, \ldots & & \mbox{Multiplier
$a=2$} \\
4,2,1,\frac12,\frac{1}4,\frac{1}8,\frac{1}{16},\frac{1}{32},\frac{1}{64},\frac{1}{128},
\ldots & & \mbox{Multiplier $a=\frac12$.} \\ 4,4\cdot 3^{1},4\cdot
3^{2},4\cdot 3^{3},4\cdot 3^{4},4\cdot 3^{5}, \ldots& &
\mbox{Multiplier $a=3$} \end{eqnarray*} Except for the first term, each
term in a geometric sequence is a constant multiple $a$ of the previous
term. Geometric sequences occur in many money computations. Observe if
\[c_k=f(k)=a^k\cdot R = (1+i)^k P\] for $k=0,1,2,3, \ldots$ then the sequence
$c_k$ forms an geometric sequence with initial element $c_0=a^0R=R.$
Instead of talking about sequences $a_1,a_2,a_3,\ldots$, we can talk
about sequences $a_m,a_{m+1},a_{m+3},\ldots$ where $m$ is any integer.
Sequences in general do not have to start with $m=1$. )
3.1 Geometric Sums
The sum of terms from a geometric sequence is called a geometric sum.
Just as there is an addition formula for adding arithmetic sequences,
there is also an addition formula for the sum of a geometric sequence.
Example: Suppose we want to compute \[S=7+7\cdot
5^1+7\cdot5^2+7\cdot5^3+7\cdot 5^4\] The five numbers in this sum form a
geometric sequence with multiplier $a=5$. We will subtract $S$ from
$5S=aS$. \begin{eqnarray*} & &\llap{a}S=\hbox{ }+7\cdot
5^1+7\cdot5^2+7\cdot5^3+ 7\cdot 5^4+7\cdot 5^5 \\
&&\underline{\llap{-}S=-7-7\cdot 5^1-7\cdot5^2-7\cdot5^3-7\cdot
5^4\qquad} \\ \hbox{Therefore } \quad &\qquad &\llap{aS-}S=-7+7\cdot
5^{5} \end{eqnarray*} This yields $(a-1)S=aS-S=7\cdot (5^5-1)=a^5-1$.
Thus \[S=7\cdot \frac{a^5-1}{a-1}= 7\cdot \frac{5^5-1}{5-1}\] The
multiplier $a$ is kept in this calculations to indicate the pattern that
is followed. Therefore \[S=7\cdot \frac{3125-1}{5-1}=7\cdot \frac{3124}{4}
=7\cdot 781=5467\]
Another Example: Compute
Compute \[S=3+3\cdot
\left(\frac23\right)^1+3\cdot\left(\frac23\right)^2+3\cdot\left(\frac23\right)^3\]
The four numbers in this sum form a geometric sequence with multiplier
$a=(\frac23)$. Again we subtract $S$ from $aS=3S$. The calculation
follows. \begin{eqnarray*} &&\llap{a}S=\hbox{ }3\cdot
\left(\frac23\right)^1+3\cdot\left(\frac23\right)^2+3\cdot\left(\frac23\right)^3+3\cdot
\left(\frac23\right)^4 \\ &&\underline{\llap{-}S=-3-3\cdot
\left(\frac23\right)^1-3\cdot\left(\frac23\right)^2-
3\cdot\left(\frac23\right)^3\qquad} \\ && \llap{ Thus \quad
aS-}S=-3+3\cdot \left(\frac23\right)^{4} \end{eqnarray*} This implies
$(a-1)S=3\cdot (\left(\frac23\right)^4-1)=3\cdot (a^4-1)$. Therefore
\[S=3\cdot \frac{a^4-1}{a-1}= 3\cdot
\frac{\left(\frac23\right)^4-1}{\frac23-1}=3\cdot
\frac{1-\left(\frac23\right)^4}{1-\frac23}\] Simplification yields
$S=\frac{65}9=7+\frac29$.
The general pattern is as follows. Suppose we want to compute
Suppose we want to compute \[S=R+R\cdot a^1+R\cdot a^2+\cdots+R\cdot
a^{m-1}+R\cdot a^m\] The five numbers in this sum form a geometric
sequence with multiplier $a$. We will subtract $S$ from $aS$.
\begin{eqnarray*} & &\llap{a}S=\hbox{ }+R\cdot a^1+R\cdot
a^2+\cdots+R\cdot a^{m}+R\cdot a^{m+1} \\
&&\underline{\llap{-}S=-R-R\cdot a^1-R\cdot a^2-\cdots-R\cdot
a^m\qquad} \\ &&\llap{\mbox{Therefore } \quad aS}-S=-R+R\cdot
a^{m+1} \end{eqnarray*} This implies $(a-1)S=R\cdot (a^{m+1}-1)$. Thus
$a\ne1$ implies \[S=R\cdot \frac{a^{m+1}-1}{a-1}\] When $m$ is large,
this formula for $S$ is quicker to calculate than the initial formula
\[S=R+R\cdot a^1+R\cdot a^2+\cdots+R\cdot a^{m-1}+R\cdot a^m\]
Exercise: Write out or think through the above argument without
the three dots $\cdots$ notation for the three cases $m=4$, $m=3$ and
$m=2$.
A Special Case. The case where $R=P\cdot a^n$ is often of interest
in money calculations. In this case, the geometric sum \[S=Pa^n+P\cdot
a^{n+1}+P\cdot a^{n+2}+\cdots+P\cdot a^{n+m-1}+P\cdot a^{n+m}\] yields
the same result as \[S=P\cdot a^n\cdot \frac{a^{m+1}-1}{a-1}\]
Remark. Geometric sums are useful in calculations involving debts,
loans, pension plans and mortgages when interest is compounded. Geometric
sums can be used to find the limiting value of repeating decimal
expansions. Geometric sums also appear in the discussion of convergence
of polynomial approximations or representation of functions and formulas.
In particular, geometric sums also appear in the polynomial approximation
of logarithms. Geometric sums play a key role in the arithmetic-based
perspective of higher mathematics.
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