Transcript Two Enrichment Exercises

The Game of Algebra
or
The Other Side of
Arithmetic
Lesson 14
by
Herbert I. Gross & Richard A. Medeiros
© 2007 Herbert I. Gross
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Two
Enrichment
Examples
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© 2007 Herbert I. Gross
Strange as it may seem, there is a connection
between how recipes are used in cooking and
how formulas are used in mathematics.
It takes a special talent to create a recipe, but
many people who do not have this talent are
still able to follow the recipe effectively.
In this context, a formula is a mathematical
“recipe” that tells us how to “mix” certain
variables to “create” other variables.
We don't have to know the derivation of a
formula in order to be able to use it. In fact,
from the very first lesson in this course, we
used formulas without first proving that they
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were correct.
© 2007 Herbert I. Gross
For example, it is not necessary to know
why the formula A = πr2 tells us how to
“create” the area (A) of a circle once we
know the radius (r) of the circle. The
formula is easy to use; namely starting
with the radius we multiply it by itself and
then multiply the result by π. However, the
actual derivation of the formula is not easy.
In fact, a rigorous derivation is usually
postponed until the study of calculus.
radius
© 2007 Herbert I. Gross
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While a person can follow a recipe without
understanding what each ingredient does
and why the ingredients are in the given
proportions, there are times when this
knowledge isn't enough.
So, in the form of enrichment, this Lesson
will attempt to provide the flavor of how
formulas are developed by looking at two
interesting examples, one of which
involves a linear relationship and the
other of which involves a relationship that
is “almost linear”.
© 2007 Herbert I. Gross
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Let's look at the linear example first.
We have already defined the standard form
of a linear relationship to be…
y = mx + b
where m and b are constants.
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Keep in mind that as written, we have no
way of knowing whether the formula refers
to an “easy” situation or a “hard” situation.
For example, at the “easy” end of the
spectrum, we have already seen that at a
cost of $4 per pound plus a charge of $5
for shipping and handling, the cost (y) in
dollars of x pounds of candy is given by
the formula…
y = 4x + 5
where m = 4 and b = 5.
© 2007 Herbert I. Gross
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We call this an “easy” situation because
most of us who were ordering the candy
would know, even without seeing the
formula, that we have to multiply the
number of pounds we buy by $4 and then
add $5 to this amount to cover the cost of
shipping and handling.
At the “harder” end of the spectrum
is the relationship between the
Fahrenheit temperature scale (F)
and the Celsius temperature (C).
© 2007 Herbert I. Gross
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Yet the method for determining the formula
is the same for all linear relationships.
First, however, let us explain why the
relationship between the Fahrenheit scale
and the Celsius scale is linear. Namely,
both scales measure the height of a
mercury column on a thermometer. When
the mercury level is at a certain height, the
thermometer doesn't know which scale
we're using.
© 2007 Herbert I. Gross
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Hence, the change in the
Fahrenheit reading has to
be proportional to the
change in the Celsius
reading. In other words, if
the temperature level in the
column rises by 1 inch, the
change in temperature is
the same even though the
readings on both scales are
different.
© 2007 Herbert I. Gross
C
F
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As a simpler illustration, think, for example,
in terms of length, if one piece of string is
twice as long as another, then no matter
what scale we use, the measurement of
one length will be twice the measurement
of the other.
1 yard = 3 feet = 36 inches
2 yards = 6 feet = 72 inches
© 2007 Herbert I. Gross
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Linear vs. Direct Proportion
There is a basic difference between
temperature scales and length scales.
Namely, while the relationship between the
perimeter of a square and the length of a
side is a direct proportion; the relationship
between Celsius and Fahrenheit is linear,
but it is not a direct proportion.
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For example, if a piece of string were to
have no length, its length would be 0
whether we measured it in inches or
whether we measured it in centimeters. On
the other hand, it is well-known that when
the temperature is 0° on the Celsius scale,
it is 32° on the Fahrenheit scale, and that
when C = 100, F = 212. At any rate, if we
elect to express F in terms of C, we now
know that the initial value of F (i.e., the
value of F when C = 0) is 32.
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That is, if we start with the standard linear
form… and replace F by 32 when C = 0, we
see that… and since m0 = 0, the equation
becomes… and if we now replace b by 32
in the formula, we get…
32F == m0
mC++bb
32 = 0 + b = b
F = mC + 32
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We also know that when C = 100, F = 212.
Hence, we may replace C by 100 and
F by 212 to obtain… Subtracting 32 from
both sides gives us… And if we now divide
both sides by 100 we obtain…
212F==100m
mC + +3232
180 = 100m
180 = m
100
© 2007 Herbert I. Gross
9
=
5
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If we now replace m in the formula by 9/5,
we obtain…
9
F = mC
/5C + 32
Note
In the linear relationship y = mx + b, notice that
b always represents the value of y when the
value of x is 0. That is, if we replace x by 0 in
the formula y = mx + b, we see that y = m0 + b,
or y = b. That's why it is helpful in starting
problems such as this to know the value of y
when x = 0. In particular, in this illustration it
was nice to know that F = 32 when C = 0.
© 2007 Herbert I. Gross
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Note
In other words, if we had started our
solution trying to use the fact that when
C = 100, F = 212, the formula F = mC + b
would have become 212 = m100 + b; and we
would have been faced with a situation in
which there was only one equation but two
unknowns. Sometimes we cannot avoid
having to deal with equations in which there
are more than one unknown. What we do in
such cases is discussed later in this course.
© 2007 Herbert I. Gross
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Defining
a Constant
Remember what a constant is. It is a
number that never changes. Hence, if
b = 32 when C = 0, then b = 32 no matter
what the value of C is. That's why we were
allowed to replace b by 32 in the formula
F = mC + b even though we only
demonstrated that b = 32
when C was equal to 0.
© 2007 Herbert I. Gross
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The Formula
If you remembered the formula
C = 5/9(F – 32), you could have obtained the
formula F = 9/5C + 32 directly by the
“undoing” method.
Namely, start with C =5/9(F – 32) and
multiply both sides by 9/5 to obtain
9/5C = F – 32; then add 32 to both sides to
obtain 9/5C + 32 = F, which by symmetry is
equivalent to F = 9/5C + 32
© 2007 Herbert I. Gross
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F or C?
Keep in mind that F = 9/5C + 32 and
C = 5/9(F – 32) are two different ways for
expressing the same relationship.
That is, we would most likely use…
F = 9/5C + 32
when we know the value of C and we
wanted to find the value of F; and we would
most likely use…
C = 5/9(F – 32)
when we know the value of F and wanted to
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fine the value of C.
© 2007 Herbert I. Gross
Development of the
Fahrenheit and Celsius
Temperature Scale
Fahrenheit was a Danish physician who
invented the first thermometer. Recognizing
that heat caused most liquids to expand and
cold caused them to contract, Fahrenheit made
a tube with a narrow opening that had a well or
reservoir at the bottom to hold the liquid. By
making the opening very narrow the liquid was
forced to move up the tube when it expanded
and down the tube when it contracted. next
© 2007 Herbert I. Gross
Fahrenheit could now measure temperature
simply by measuring the height of the
liquid level in the tube. To do this, he had
to develop a scale.
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So on a particularly cold morning in
Copenhagen, he left the thermometer
outside his window, and he marked the
liquid level as 0°F (actually, he simply called
it 0° because at the time there were no other
temperature scales). He then took his body
temperature and marked the liquid level as
100°F (which means that either he had a
slight fever that day or else he didn't
measure too accurately. Otherwise, normal
body temperature today would be 100°F
rather than 98.6°F)
© 2007 Herbert I. Gross
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Celsius later raised the issue of who cared
about either the temperature in
Copenhagen on one particular cold
morning or Fahrenheit's body temperature
on a particular day? Rather he wanted to
pick a scale that would be more readily
recognizable to any scientist.
That is, he wanted to use readings of 0°and
100° to represent temperatures that any
scientist could reproduce at any time in any
laboratory.
© 2007 Herbert I. Gross
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Accordingly he used 0°C to represent the
temperature at which water froze and he
used 100°C to represent the temperature at
which water boiled.
Using these two scales, it turned out that…
-- when the Celsius reading was 0°, the
Fahrenheit reading was 32°;
-- and when the Celsius reading was 100°,
the Fahrenheit reading was 212°.
© 2007 Herbert I. Gross
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In other words, there are
180º F per 100º C.
Written as a fraction
180º F = 9ºF
100º C 5ºC
212º F
180º F
32º F
100º C
100º C
0º C
This explains why
m = 9/5
© 2007 Herbert I. Gross
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Rate of Change
Let's now turn our attention to an
interesting relationship that is
“almost linear”. More specifically there
are times when the rate of change is not
constant, but the rate of change of the
rate of change is constant. “the rate of
change of the rate of change” might
sound a bit like a tongue twister, but it is
something that we deal with quite often.
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For example, the speed of an object is
already a rate of change. Namely, it is the
rate of change of the distance the object
travels with respect to the time it took to
travel this distance.
The acceleration of the object is the rate of
change of its speed with respect to time.
Thus, acceleration may be viewed as a rate
of change of a rate of change.
For example, if the speed is measured in feet
per second, the acceleration could be
measured in feet per second per second or
feet per second per minute, etc.
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© 2007 Herbert I. Gross
So, for example, suppose the acceleration
of an object is 2 feet per second per
second. In other words, suppose that the
speed of the object increases by 2 feet per
second every second. If the object starts
from a stopped position, at the end of 1
second its speed is 2 feet per second; at
the end of 2 seconds its speed is 4 feet per
second; at the end of 3 seconds its speed
is 6 feet per second etc. Thus, the speed of
the object is not constant, but its rate of
change (acceleration) is constant.
© 2007 Herbert I. Gross
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Another example is the rate of change of
the area (A) of a square with respect to the
length (L) of one of its sides. The formula
is A = L2. In the chart below the third
column represents the rate of change of A
with respect to L when L increases by 1;
and the fourth column represents the rate
of change of the third column with respect
to L when L increases by 1.
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L
1
In terms of a
chart…
A
1
Rate of
Change
Rate of Change of
the Rate of Change
3
2
4
2
5
3
9
2
7
A = L2
4
16
2
9
5
25
2
11
6
36
2
13
© 2007 Herbert I. Gross
7
49
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There is an interesting pattern that seems
to be indicated in the previous chart.
For example:
1 + 3 = 4 = 22
1 + 3 + 5 = 9 = 32
1 + 3 + 5 + 7 = 16 = 42
1 + 3 + 5 + 7 + 9 = 25 = 52
1 + 3 + 5 + 7 + 9 + 11 = 36 = 62
1 + 3 + 5 + 7 + 9 + 11 + 13 = 49 = 72
© 2007 Herbert I. Gross
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To see the last equality more visually, we
may think of 49 (that is, 72) as the area of a
square, each of whose side is 7 units. In
the diagram below notice that the shaded
L-shaped regions represent the odd
numbers, 1, 3, 5, 7, 9,11, and 13.
1
3
5
7
9
11
© 2007 Herbert I. Gross
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Another example is to suppose that we want
to form the sum of consecutive integers
starting with 1. For example, if we let Sn
denote the sum of the first n integers
starting with 1, we see that…
S1 = 1
S2 = 1 + 2 = 3
S3 = 1 + 2 + 3 = 6
S4 = 1 + 2 + 3 + 4 = 10
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S1 = 1
-- In forming S2 we added 2 to the previous
sum.
S2 = 1 + 2 = 3
-- In forming S3 we added 3 to the previous
sum.
S =1+2+3=6
3
-- In forming S4 we added 4 to the previous
sum.
S = 1 + 2 + 3 + 4 = 10
4
-- In general in forming Sn we add n to the
previous sum.
-- Thus, each term in the sum is 1 more
than the previous term.
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© 2007 Herbert I. Gross
As the number of terms in the sum
increases the process of adding term by
term becomes more and more tedious.
For example, suppose we want to compute
the value of S200; that is,
1 + 2 + 3 + 4 + …+ 198 + 199 + 200.
Certainly, as we've already seen it wasn't
tedious to compute, say S4, but imagine
trying to keep on going this way until we got
to 200. More specifically, the main problem
with this approach is that in order to find
the sum of the first 200 whole numbers, we
first have to find the sum of the first 199, etc.
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© 2007 Herbert I. Gross
However, there is a pattern to the terms that
gives us a way to form the sum without
having to add the terms successively. The
pattern is based on the fact that the
difference between successive terms is
always the same. For example, the logic we
are going to use would have been the same
if we wanted to find the sum of the first 200
even positive integers; that is…
2 + 4 + 6 + 8 + 10 + ....... + 400
© 2007 Herbert I. Gross
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To see the “trick” let's look at a simpler case
such as S5
1+2+3+4+5
-- Notice that if we start at 1 and look at
the terms from left to right, each term is 1
more than the previous one; but if we start
at 5 and look at the terms from right to left
each term is 1 less than the previous one.
© 2007 Herbert I. Gross
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Since the increase of 1 balances a decrease of
1, it means that if we start at both ends and add
the terms in pairs, each of the pairs will have
the same sum. This might be more visual if we
represented the above in the form…
+
S5 = 1 + 2 + 3 + 4 + 5
S5 = 5 + 4 + 3 + 2 + 1
2S5 = 6 + 6 + 6 + 6 + 6 = 5 × 6
2
2
5×3
And if S
we
5 =now divide both sides of the
equation by 2, we see that…
© 2007 Herbert I. Gross
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The idea is the same whether there are 5 terms
or 200 terms or 2 million terms! Namely, when
we add any number of consecutive whole
numbers starting with 1, the sum of the first
term and the last term will be 1 more than the
last term (because no matter what the last term
is, the first term is always 1); and if we pair the
terms starting from both ends, each of the pairs
will have the same sum.
In the case of adding the first 200 whole
numbers, the sum of the first term and the last
term is 201; and since there are 200 terms
there are 100 pairs; so the sum will be
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201 × 100, or 20,100.
© 2007 Herbert I. Gross
More generally, if we let Sn stand for the
sum of the first “n” positive whole
numbers…
Sn = 1 + 2 + 3 + ........... + n,
Then…
© 2007 Herbert I. Gross
Sn = (n + 1) • n
2
= (n + 1) • n
1
2
= (n + 1)n
2
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A Geometric Interpretation
To visualize why, for example, S5 = 5 × 6,
2
look at the figures below…
Think of square tiles
where each tile represents
1 unit of area.
The shaded tiles in the above diagram
represent 15 units of area, which may also
be viewed as 1 + 2 + 3 + 4 + 5 = S5 next
© 2007 Herbert I. Gross
In a similar way the shaded tiles below also
indicate 15 units of area as well as the sum
1 + 2 + 3 + 4 + 5 = S5
© 2007 Herbert I. Gross
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We may put these
two diagrams
together to form the
rectangle below.
On the one hand the area of the rectangle
is 6 × 5 or 30 units, and on the other hand it
is 2 × S5. Hence S5 = 6 × 5 .
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2
© 2007 Herbert I. Gross
Another Application
Suppose there are 6 people in a room
(whom we shall name simply by
1, 2, 3, 4, 5, and 6),
and they all exchange handshakes.
We can compute the number of
handshakes in 2 ways.
© 2007 Herbert I. Gross
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One Way
Each of the 6 people shakes hands with
each of the other 5 people. Thus it would
seem that there were 6 × 5 or 30 hand
shakes. However, this method counted
each handshake twice; for example, when
1 shakes hands with 2 it's the same
handshake as when 2 shakes hands with 1.
Therefore the total number of handshakes
is 6 × 5.
2
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© 2007 Herbert I. Gross
The Second Way
Another way to think of it is to visualize
that the room is initially empty and the
people come in one at a time; say in the
order of 1, 2, 3, 4, 5 and 6.
1 comes in first and there is no one to shake hands with.
2 shakes hands with 1; which we will denote by (2,1)
3 shakes hands with 1 and 2; that is (3,1) (3,2)
4 shakes hands with 1, 2 and 3; that is (4,1) (4,2) (4,3)
5 shakes hands with 1, 2, 3 and 4; that is (5,1) (5,2) (5,3) (5,4)
6 shakes hands with 1, 2, 3, 4, 5 and 6; that is
(6,1) (6,2) (6,3) (6,4) (6,5)
© 2007 Herbert I. Gross
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Handshake Summary
The number of handshakes on the one
hand is given by…
6 × 5,
2
and on the other hand by…
1 + 2 + 3 + 4 + 5.
Hence, 1 + 2 + 3 + 4 + 5 = 6 × 5 = 5 × 6
2
2
© 2007 Herbert I. Gross
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A Note on Arithmetic Series
The sum 1 + 2 + 3 + ... + n is a special
case of what mathematicians refer to as
an arithmetic series.
Definition
An arithmetic series is any sum in which
the difference between two consecutive
terms is constant.
© 2007 Herbert I. Gross
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The first term can be any number we
choose and we then pick the remaining
terms so that the difference between
consecutive terms is always the same.
(The sum 1 + 2 + 3 + 4 + 5 + 6 + ... + n is
the special case of an arithmetic sum in
which the first term is 1 and the
difference between consecutive terms in
the sum is also 1).
© 2007 Herbert I. Gross
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For example, suppose we choose the first
term to be 6 and we then choose each new
term to be 4 more than the previous one.
Thus, the arithmetic series would look like…
T1 = 6
T2 = 6 +10
T3 = 6 + 10 +
T4 = 6 + 10 +
T5 = 6 + 10 +
T6 = 6 + 10 +
14
14 + 18
14 + 18 + 22
14 + 18 + 22 + 26
T7 = 6 + 10 + 14 + 18 + 22 + 26 + 30
© 2007 Herbert I. Gross
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Point
The important point is that as we move
from left to right each term is 4 more
than the previous term; and this is
equivalent to saying that as we move
from right to left each term is 4 less than
the previous term.
© 2007 Herbert I. Gross
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So, for example, one way to compute, say,
T6 is to mimic what we did previously.
Namely, we write the sum twice, once from
left to right and once from right to left. In
this way we obtain…
T6 = 6 + 10 + 14 + 18 + 22 + 26
+ T6 = 26 + 22 + 18 + 14 + 10 + 6
2T6 = 32 + 32 + 32 + 32 + 32 + 32 = 6 × 32
2
2
3 × 32of the
And if
T6we
= now divide both sides
equation by 2, we see that…
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= 96
© 2007 Herbert I. Gross
Notice that to find the sum T6…
-- First, we added the first term (6) and
the last term (26).
-- Then, we multiplied this sum by the
number of terms in the
sum (6).
-- Finally, we divided the above answer
by 2.
© 2007 Herbert I. Gross
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The above procedure would work for
finding the sum of any arithmetic series.
Namely…
Step 1: Add the first term (F) and the last
term (L).
Step 2: Multiply the result in step one by
the number of terms in the sum (n).
Step 3: Divide the answer in Step 2 by 2.
© 2007 Herbert I. Gross
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In terms of a formula…
Tn = n(F + L)
2
In the special case,
Tn = 1 + 2 + 3 + 4 + 5 + ... + n,
F = 1 and L = n
© 2007 Herbert I. Gross
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An interesting closing example might be to
revisit our earlier example concerning the
fact that the sum of consecutive odd
numbers, starting with 1, is always a perfect
square. By way of review…
1 = 12
1 + 3 = 4 = 22
1 + 3 + 5 = 9 = 32
In general, the sum of the first n consecutive
odd numbers is n2.
Thus for example, the sum of the first 30
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odd numbers is 302 or 900.
© 2007 Herbert I. Gross
However, suppose we hadn't noticed
this fact and wanted to find the sum of
the first 30 odd numbers; that is:
1 + 3 + 5 + 7 + 9 + ... + 59.
The above is an arithmetic series because
each term always exceeds the previous
term by 2.
© 2007 Herbert I. Gross
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Applying the Formula to the
Sum of the First 30 Odd Numbers.
-- The sum of the first term (1) and the
30th term (59) is 60.
-- The number of terms is 30 and
30 × 60 = 1,800.
-- 1,800 ÷ 2 = 900 = 302
© 2007 Herbert I. Gross
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