Download focus on problem solving 10

FOCUS ON PROBLEM SOLVING 10
The solutions to many of the problems of mathematics involve finding
patterns. The algebraic formulas we have found in this book are
compact ways of describing a pattern. For example, the familiar
equation (a + b) 2 = a 2 + 2ab + b 2 gives the pattern for squaring the sum
of two numbers. Another example we have encountered is the pattern
for the sum of the first n odd numbers:
1+ 3 + 5 +
+ (2n − 1) = n 2
How do we discover patterns? In many cases, a good way to start is to
experiment with the problem at hand. How can we be sure a pattern
always holds? One way is to use mathematical induction, but there are
other ways. In the example we give here we find a pattern for adding
the cubes of the first n natural numbers. We then find a geometrical
interpretation of this sum that shows the pattern always holds. This
startling connection between number patterns and geometrical patterns
is credited to the 11th century mathematician Abu Bekr Mohammed
ibn Alhusain Al Karchi.
The Gnomons of Al Karchi
We prove the beautiful formula
13 + 23 + 33 +
+ n3 = (1 + 2 + 3 +
+ n) 2
But first, here is how this formula was discovered. The sum of the first
n natural numbers is called a triangular number. The first few
triangular numbers are 1, 1 + 2 = 3 , 1 + 2 + 3 = 6 , 1 + 2 + 3 + 4 = 10 , and
in general.
1+ 2 + 3 +
+n=
n(n + 1)
2
The name “triangular” comes from the following pictures:
1
The sequence of triangular numbers
1,3, 6,10,15,…
is so well known that any mathematician would instantly recognize it.
Now let’s look at the sums of the cubes:
13 = 1
13 + 23 = 9
13 + 23 + 33 = 36
13 + 23 + 33 + 43 = 100
13 + 23 + 33 + 43 + 53 = 225
We get the sequence
1,9,36,100, 225,…
We notice that the numbers in this sequence are the squares of the
triangular numbers. It appears that the sum of the cubes of the first n
natural numbers equals the square of the sum of the first n natural
numbers.
To show that this pattern always holds, Al Karchi sketches the
following diagram.
2
Each region Gn in the shape of an inverted L is called a “gonomon.”
Let us find the area of the gnomon G4 . From the figure in the margin
we see that the area is
42 + 2[4 × (1 + 2 + 3)] = 64
In general, the areas of the gnomons G1 , G2 , G3 , G4 , G5 ,… are
1,8, 27, 64,125,… . These are the cubes of the natural numbers. The
pattern persists here also, since the area of the nth gnomon Gn is n3 ,
as the following calculation shows:
n 2 + 2[n × (1 + 2 +
(n − 1)n
2
2
3
2
= n +n −n
+ (n − 1))] = n 2 + 2n
= n3
Now comes Al Karchi’s punch line. The first n gnomons form a square
of side 1 + 2 + 3 + + n and so the sum of the areas of these gnomons
equals the area of the square. That is, 13 + 23 + + n3 = (1 + 2 + + n) 2 .
PROBLEMS
1. Use the diagram shown to find and prove a formula for the sum of
the first n odd numbers.
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2.
Find formulas for the following products.
⎛
⎝
(a) ⎜ 1 −
(a)
3.
1 ⎞⎛ 1 ⎞⎛ 1 ⎞
⎟ ⎜1 − ⎟ ⎜1 − ⎟
2 ⎠⎝ 3 ⎠⎝ 4 ⎠
1⎞
⎛ 1 ⎞⎛ 1 ⎞⎛
⎜1 − ⎟ ⎜1 − ⎟ ⎜1 − ⎟
⎝ 4 ⎠ ⎝ 9 ⎠ ⎝ 16 ⎠
⎛ 1⎞
⎜1 − ⎟
⎝ n⎠
1 ⎞
⎛
⎜1 − 2 ⎟
⎝ n ⎠
Prove that
1
1
1
+
+
+
sin 2 sin 4 sin 8
+
1
= cot1 − cot 2n
sin 2n
4. Prove that
n5 n 4 n3 n
+ + −
5 2 3 30
is an integer for all natural numbers n.
5. Find a formula for the sums
S n = 1⋅1! + 2 ⋅ 2! + 3 ⋅ 3! +
+ n ⋅ n!
and prove that your formula holds for all n.
6. Find a formula for the sums
Sn =
1 2 3
+ + +
2! 3! 4!
+
n
(n + 1)!
and prove that your formula holds for all n.
7. Find the following digits.
(a) If S = 100! , how many of the digits at the end of the number
S
are zeros?
(a) If
S = 1!+ 2!+ 3!+
+ 99!, what is the last digit in the value of
S.
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