Paul Abbott​
School of Physics, Mathematics and Computing
The University of Western Australia
​paul.c.abbott@uwa.edu.au
In this Chat-Enabled notebook I revisit questions asked in the paper Empowering mathematics education through programming.

The sum of even natural numbers

In my view, I think the more important question to ask is what is the sum of
n
terms, something that Python and Geogebra will likely struggle with.
Sum of the first n even natural numbers
The sum of the first
n
even natural numbers can be calculated using a simple formula. The first
n
even natural numbers are
2,4,6,…,2n
. ​The sum
S
of these numbers can be given by: ​
S2+4+6+⋯+2n
​This can be factored as: ​
S2(1+2+3+⋯+n)
​The sum of the first
n
natural numbers
1+2+3+⋯+n
is given by the formula: ​
n(n+1)
2
​Therefore, the sum of the first
n
even natural numbers is: ​
S2
n(n+1)
2
n(n+1)
​So, the sum of the first
n
even natural numbers is
n(n+1)
.​Let's verify this with an example. For
n5
: ​
S5(5+1)5630
​And the first 5 even natural numbers are 2, 4, 6, 8, 10 , whose sum is indeed 30 .
Give me Mathematica code for computing this sum.
Sure! Here is the Mathematica code to compute the sum of the first
n
even natural numbers: ​
SumOfEvenNaturalNumbers[n_]:=n(n+1)
Insert and evaluate
Insert
Copy
​You can use this function to compute the sum for any
n
. For example, to compute the sum of the first 5 even natural numbers: ​
SumOfEvenNaturalNumbers[5]
Insert and evaluate
Insert
Copy
​This will return 30, as expected.
1. How many terms are needed to get a sum larger than 10,000?
2. Can we get the sum 100?
Let's address each question one by one. ​
1. How many terms are needed to get a sum larger than 10,000?
​We need to solve for
n
in the inequality: ​
n(n+1)>10,000
​To find the smallest integer
n
that satisfies this inequality, we can solve the quadratic equation: ​
2
n
+n-10,0000
​Using the quadratic formula: ​
n
-b±
2
b
-4ac
2a
​where
a1
,
b1
, and
c-10,000
, we get: ​
n
-1±
1+40,000
2

-1±
40,001
2
​Since
n
must be positive, we take the positive root: ​
n
-1+
40,001
2
​Let's compute this value. ​
n=(-1+Sqrt[40001])/2​​Ceiling[n]
Insert and evaluate
Insert
Copy
​​
2. Can we get the sum of 100?
​We need to check if there exists an integer
n
such that: ​
n(n+1)100
​This is a quadratic equation: ​
2
n
+n-1000
​Again, we use the quadratic formula: ​
n
-1±
1+400
2

-1±
401
2
​Let's compute this value. ​
n=(-1+Sqrt[401])/2
Insert and evaluate
Insert
Copy
​If
n
is an integer, then it is possible to get the sum of 100 using the first
n
even natural numbers. Otherwise, it is not possible.​Let's calculate these results. ​
nLargerThan10000=Ceiling[(-1+Sqrt[40001])/2]​​nSum100=(-1+Sqrt[401])/2
Insert and evaluate
Insert
Copy
​With these calculations, we can determine the answers to both questions.

Code Evaluation

Mouse-over the code, select, and evaluate:
In[]:=
nLargerThan10000=Ceiling[(-1+Sqrt[40001])/2]
Out[]=
100
In[]:=
nSum100=(-1+Sqrt[401])/2
Out[]=
1
2
(-1+
401
)

Sum of all four digit numbers divisible by three

Sum of all four digit numbers divisible by three.
To find the sum of all four-digit numbers divisible by 3, we can use the formula for the sum of an arithmetic series. ​The smallest four-digit number is 1000, and the largest is 9999. We need to find the four-digit numbers within this range that are divisible by 3.​1. Identify the smallest and largest four-digit numbers divisible by 3:​ • The smallest four-digit number divisible by 3 is the smallest number greater than or equal to 1000 that is divisible by 3. Since
1000÷3333
with a remainder of 1, the smallest four-digit number divisible by 3 is 1002. • The largest four-digit number divisible by 3 is the largest number less than or equal to 9999 that is divisible by 3. Since
9999÷33333
with no remainder, the largest four-digit number divisible by 3 is 9999.2. Calculate the number of terms in the sequence:​ • The sequence of four-digit numbers divisible by 3 from 1002 to 9999 forms an arithmetic sequence with the first term
a1002
, the common difference
d3
, and the last term
l9999
. • The number of terms
n
can be found using the formula for the
n
-th term of an arithmetic sequence:
a
n
a+(n-1)d
. Setting
a
n
9999
, we get:
99991002+(n-1)·3
​ Solving for
n
:
9999-1002(n-1)·3
​
8997(n-1)·3
​
n-1
8997
3
​
n-12999
​
n3000
​3. Calculate the sum of the sequence:​ • The sum
S
of an arithmetic sequence can be calculated using the formula:
S
n
2
(a+l)
​ Substituting the values we found:
S
3000
2
(1002+9999)
​
S1500·11001
​
S16501500
​So, the sum of all four-digit numbers divisible by 3 is 16,501,500.

Summarize all three-digit numbers divisible by seven

What is the sum of all three digit numbers divisible by seven.

Triangle with unknown angles

A triangle ΔABC has three unknown angles called a, b, and c with the following relations. Angle a is 12° larger than angle c. Angle b is four times larger than angle c. Determine the size of the angles.
A triangle ΔABC has three unknown angles a, b, and c with the relations: Angle a is 5° larger than angle c. Angle b is 50° larger than angle c+a. Determine the size of the angles.

Comments

Regarding “The programming seemed to be much easier for most of the students than I had imagined”, with Chat-Enabled notebooks, this has become even easier.
“Of course, some students disliked entering the mathematics into the environments of GeoGebra, Python, and Wolfram Alpha. The[y] were probably considering mathematics to be something that should be done with paper and pencil and thus considering that letting GeoGebra, Python, and Wolfram Alpha solve the problem was cheating.”
In my view it is not cheating. The point is that the entire approach and goal of math(s) teaching needs to be re-thought because of what current technology is capable of.
Bruno Buchberger’s White-Box/Black-Box principle addresses what should be done with paper and pencil.

Buchberger (1990)

Proposed White-Box/Black-Box principle (WBBB) to describe computer-assisted learning:
In the stage where area X is new to the students, the use of a symbolic software system realizing the algorithm of area X as black boxes would be a disaster. Students have to study the area thoroughly, that is they should study problems, basic concepts theorems, proofs, algorithms based on the theorems, examples, and hand calculations.
In the stage where area X has been thoroughly studied, when hand calculation for simple examples become routine, and hand calculations for complex problems becomes intractable, students should be allowed and encouraged to use the respective algorithms available in symbolic software systems.

Kutzler (2000)

Modified the WBBB principle: students who do not master certain topics can use computer-assisted learning as a scaffolding to help understand higher level topics.

Drijvers (2000)

Proposed inverting the sequence—Black-Box/White-Box—using computer-assisted learning as a generator of examples and as an explorative tool that confronts the student with new situations. The explorative Black-Box phase can lead to (directed) discoveries. In the White-Box phase the results of the explorations are examined and can lead to the development of new concepts.
References
1
Buchberger, Bruno. “Should Students Learn Integration Rules?” SIGSAM Bull. 24 (1990): 10-17. https://doi.org/10.1145/382276.1095228
2
Kutzler, Bernhard. “The Algebraic Calculator as a Pedagogical Tool for Teaching Mathematics.” International Journal of Computer Algebra in Mathematics Education, 7(1), 5-23 (2000). https://www.learntechlib.org/p/92408/
3
Drijvers, Paul. “Students encountering obstacles using a CAS.” International Journal of Computers for Mathematical Learning 5, 189–209 (2000). https://doi.org/10.1023/A:1009825629417

CITE THIS NOTEBOOK

Empowering Mathematics education through programming​
by Paul Abbott​
Wolfram Community, STAFF PICKS, October 1, 2024
​https://community.wolfram.com/groups/-/m/t/3288988