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The goal of homeworks is practice on the topics covered in the text and
in class. If you’re unsure how to tackle one problem, look at the problems
nearby or at examples. One may be more clear to you and help you with
the assigned problem.
I have office hours now. Monday and Wednesday 1.30pm to 2.30pm (or
possibly later) in the Math Lab down the hall.
Be sure to read the entire problem. Many submissions contained only
partial answers even when it was clear you understood the mechanism.
With problems involving large numbers, expect most calculators and computer
software to break. Try to check results using properties of the input
numbers. For example the product of two numbers with units digit 1 also
has units digit 1. Or that the product of two d
digit numbers has either 2d
or 2d−1
digits. (Think about long-hand multiplication to find these and other
properties.)
If there are questions about which problems were assigned or what the
problem is asking, contact me even if it’s the night before the homework
is due! I may not respond instantly, but it’s worth a shot.
Because there was apparent confusion over which problems were assigned,
I will start providing the homework on a separate page as well as directly
in the notes.
In general, writing out steps cushions the blow if the result is incorrect.
And writing out reasons helps even more. If your homework must be
late, reasoning in your own style and words shows you did not just
copy solutions. This class is as much about the method of thinking andcommunicating as it is about the final results!
Remember that homework is one 20% chunk. But there will be 14 or 15
assignments. Each is at most …
And if there are 10-20 problems per assignment, then each assignment is at
most…
This is another reason why homeworks are frequent. The impact of each
assignment is a little less when there are many.
A list of numbers as in Section 1.1 (2, 9, 16, 23, 30) does not mean anything on its
own. The context before this example implies that one should look for an arithmetic
relationship.
The “trick” is that a premise is withheld. As in poorly written mystery novels, crucial
information is not available.
All reasoning is based on premises (hypotheses, suppositions, etc.) wether implicit or
explicit. “Trick” questions like Section 1.1’s example rely on misleading you into
using an incorrect implicit premise.
The formula provided in the text is of order 4, or in other words the highest power of the
argument n
is n4.
(Another phrase for this is that the formula is quartic.) We expect to need 4 columns
to the right of the original sequence (1, 2, 4, 8, 16, 31) to reach an arithmetic
sequence.
points
regions
Δ(1)
Δ(2)
Δ(3)
Δ(4)
1
1
2
2
1
3
4
2
1
4
8
4
2
1
5
16
8
4
2
1
6
31
15
7
3
1
7
57
26
11
4
1
8
99
42
16
5
1
The formula provided is
R(n)=124n4−6n3+23n2−18n+24
One can compute this directly with any method to verify that the answer is
99.
One convenient way to rewrite a polynomial for evaluation is Horner’s rule. Horner’s
rule applies the distributive property of multiplication over addition to pull factors of
n out of
subexpressions. This rule not only is faster when using a calculator, it also incurs fewer rounding
errors when n
is not an integer.
The common property is that squaring a number with alternating 1 and 0
digits
With these short sequences, the zeros only serve to make the pattern more obvious. Note
that 112=121
and 1112=12321.
This pattern will break after the central digit is 9. Why?
Note that computing 1010101012with common desktop computers may produce10203040504030200. The last digit falls off the end of how computers representfloating-point numbers. Computing in integers on “32-bit” computers may produce 28if the calculation wraps around the 32-bit boundary.
This is one reason why looking for patterns and developing a number sense isimportant. Errors in calculated results depend on the method used for calculation.Most programs or devices do not explain their methods, so recognizing patterns andother properties are important to prevent being misled.
There are two clear ways to extend the formula
S(n)=n(n+1)∕2 into a formula
for the sum 2+4+6+⋯+2n.
One is to recognize that 2+4+6+⋯+2n=2(1+2+3+⋯+n)=2S(n)=n(n+1).
Rephrasing the original problem using summation notation, we have used
∑ni=12i=2∑ni=1i=n(n+1).
Another is to consider the sum 2+4+6+⋯+2n=(1+1)+(2+2)+(3+3)+⋯+(n+n)=(1+2+3+⋯+n)+(1+2+3+⋯+n)=S(n)+S(n)=2S(n)=n(n+1).
In summation notation, ∑ni=12i=∑ni=1(i+i)=∑ni=1i+∑ni=1i=n(n+1).
Each column (or row) of the blocked triangles represents the integers 1, 2, 3, and 4 by
the number of blocks in the column (or row). The total number of blocks in each
triangle is 1+2+3+4.
When flipped and combined, the total number blocks is the sum of the two triangles, or
2⋅(1+2+3+4). The combined figure is
a rectangle consisting of 4⋅5
blocks. So 2(1+2+3+4)=4⋅5,
or 1+2+3+4=(4⋅5)∕2.
A better diagram would replace the left-most arrow with an addition operator(+).
Drawing out the dots demonstrates the solution directly.
Or use the result of Problem 16. Note that the first sum,
∑ni=1i, is the
nth triangular number and
that the second sum, {\mathop{\mathop{∑
}}\nolimits }_{i=1}^{n-1}i,
is the (n - 1)th
triangular number. Thus Problem 16 demonstrated that the sum of two consecutive
triangular numbers is a square.
The first pattern to observe is they are all of the form
p(n)∕2, where
p(n) is some polynomial
of n. The next pattern
is that the numerator p(n) = n(a ⋅ n - b)
for integers a
and b. Then
both a
and b
increase by one when adding a side.
Using these patterns, the nth
nonagonal number is
N(n) = {n(7n - 5)\over
2} .
Substituting 6, N(6) = 6(7 ⋅ 6 - 5)∕2 = 3(42 - 5) = 111,
adding more evidence to the conjecture.