Prove Maasei Hoshev: Engineering Student Minor in Math Needs Help

[JoshC]

I am taking a course in the history of modern math. Note, I am an engineering student minoring in math. Therefore, I am not that up to speed on induction proofs. I have been working on a problem in my book (A History of Mathematics by Victor Katz), and really don't know how to procede. Any help would be greatly appreciated.
Problem:
------------------------------
Prove Proposition 32 of the Maasei hoshev (by Levi Ben Gerson):

1+(1+2)+(1+2+3)+...+(1+2+...+n)

={1^2+3^2+...+n^2 n odd;
{2^2+4^2+...+n^2 n even
------------------------------
Thank you for your assistance

 

[JoshC]

Nevermind. I dropped the course. Thanks for the help.

 

[tacman]

in advance.

First of all, it is great that you are seeking help and reaching out for assistance. This shows that you are determined to understand the material and willing to put in the effort to improve your understanding of induction proofs.

To prove Proposition 32 of the Maasei hoshev, we will use mathematical induction. This is a common technique used in mathematics to prove statements that involve a variable, such as n in this case. The steps of mathematical induction are as follows:

1. Base case: We start by proving the statement for the smallest value of n. In this case, n=1. We can see that the left side of the equation becomes 1, and the right side becomes 1^2, which are equal. Therefore, the statement holds true for n=1.

2. Inductive hypothesis: Assume that the statement is true for some arbitrary value of n, denoted as k. In other words, assume that the equation holds true for the sum of the first k terms.

3. Inductive step: Now, we need to prove that the statement is also true for the next value of n, which is k+1. This is where the induction proof comes into play. We will use our assumption from the previous step to prove that the statement holds true for n=k+1.

Let's consider the case when n is odd. In this case, the left side of the equation becomes:

1+(1+2)+(1+2+3)+...+(1+2+...+k)+(1+2+...+(k+1))

Using our assumption from the inductive hypothesis, we can rewrite the right side of the equation as:

1^2+3^2+...+k^2+(k+1)^2

Now, we can simplify the left side of the equation as:

1+2+3+...+k+(k+1)

Using the formula for the sum of the first n natural numbers, this becomes:

(k+1)(k+2)/2+(k+1)

Multiplying and simplifying, we get:

(k+1)^2+(k+1)

Which is equal to the right side of the equation, proving the statement for n=k+1 when n is odd.

Similarly, for the case when n is even, we can follow the same steps and use the formula for the sum of the first n even numbers to prove the statement.

Therefore, by using