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Answer: General Form of Sequence

Saturday, 1 of March , 2008 at 9:59 pm

Question: The solution to x = \sqrt{10y+1} where x,y \in \mathbb N yields the following sequence:

8,~12,~36,~44,~84,~96,~152,~168,~240~\ldots

Find the general form for this sequence.

Answer: When analyzing this sequence, the first thing I tried was comparing the difference between each number in the sequence.

Lets investigate this new sequence.

4,~24,~8,~40,~12,~56,~16,~72~\ldots

It appears the odd terms are related to each other and so are the even terms. Lets break the odd terms and even terms into two different sequences.

odd~terms:~4,~8,~12,~16,\ldots
even~terms:~24,~40,~56,~72,\ldots

As you can see, the odd terms has the form:

\phi_n = 4n

The even terms, likewise, can be seen to have the form:

\phi_n = \phi_{n-1} + 16 or
\phi_n = 16n + 8

If we merge these results, we can get pattern for the following sequence

4,~24,~8,~40,~12,~56,~16,~72~\ldots

which is

\phi_n = \begin{cases}4(\frac{n+1}{2}) & odd~n \\ 16(\frac{n}{2}) +8 & even~n\end{cases}
\phi_n = \begin{cases}2(n+1) & odd~n \\ 8(n+1) & even~n\end{cases}

We can then use this result to find the solution to

8,~12,~36,~44,~84,~96,~152,~168,~240~\ldots

which would be

\psi_n = \psi_{n-1} + \phi_n
\psi_n= \psi_{n-1} + \begin{cases}2(n+1) & odd~n \\ 8(n+1) & even~n\end{cases}

where \psi_0 = 8.

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Category: Algebra, Answers

Question: General Form of Sequence

Sunday, 24 of February , 2008 at 12:34 pm

Question: The solution to x = \sqrt{10y+1} where x,y \in \mathbb N yields the following sequence:

8, 12, 36, 44, 84, 96, 152, 168, 240…

Find the general form for this sequence.

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Category: Algebra, Questions

Answer: Cubes of three Consecutive Integers

Saturday, 9 of February , 2008 at 7:05 pm

Question: Show that the sum of the cubes of any three consecutive integers is divisible by 9.

Answer: In order to solve this problem, I am going to define the questions a little more formally.

Prove that there \exists a k \in \mathbb Z where f(n) = 9k given

f(n) = (n)^3 + (n+1)^3 + (n+2)^3 and n \in \mathbb Z

I will be solving this problem using induction.

Base Case:
First we need show that the problem is true for the base case (n=0)

f(0) = 0^3 + (0 + 1)^3 + (0 + 2)^3 = 0 + 1 + 8 = 9

Therefore, there \exists a k \in \mathbb Z (which would be k = 1) where f(0) = 9k.

Induction Step (for positive values):
In order to prove that the problem holds for n \in \mathbb N, we need to show:

If f(n) = 9k ~~ \Longrightarrow ~~ \exists~a~j \in \mathbb Z where f(n+1) = 9j where n,k \in \mathbb Z

In order to prove this statement, we need to reduce f(n+1).

f(n+1) = (n+1)^3 + (n+2)^3 + (n+3)^3
= n^3 + (n+1)^3 + (n+2)^3 + (n+3)^3 - n^3
= f(n) + (n+3)^3 - n^3
= f(n) + (n+3)(n+3)(n+3) - n^3
= f(n) + (n^3 + 9n^2 +27n + 27) - n^3
= f(n) + 9(n^2 +3n + 3)
= 9k + 9(n^2 +3n + 3)
= 9(n^2 +3n + 3 + k)

Since (n^2 +3n + 3 + k) \in \mathbb Z, we have shown if f(n) = 9k it implies that \exists~a~j \in \mathbb Z where f(n+1) = 9j.

Induction Step (for negative values):
Since we already proved by induction that the problem holds for all natural numbers, all we need to do to extend this proof to all integers is to show that:

If f(n) = 9k ~~ \Longrightarrow ~~ \exists~a~j \in \mathbb Z where f(n-1) = 9j where n,k \in \mathbb Z

In order to prove this statement, we need to reduce f(n-1).

f(n-1) = (n-1)^3 + n^3 + (n+1)^3
= (n-1)^3 - (n+2)^3 + n^3 + (n+1)^3 + (n+2)^3
= f(n) + (n-1)^3 - (n+2)^3
= f(n) + (n^3 - 3n^2 + 3n - 1) + (n^3 + 6n^2 + 12n + 8)
= f(n) + (-9n^2 - 9n - 9)
= f(n) + 9(-n^2 - n - 1)
= 9k + 9(-n^2 - n - 1)
= 9(-n^2 - n - 1 + k)

Since (-n^2 - n - 1 + k) \in \mathbb Z, we have shown if f(n) = 9k it implies that \exists~a~j \in \mathbb Z where f(n-1) = 9j.

Therefore, we have proven by induction that the sum of the cubes of any three consecutive integers is divisible by 9.

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Category: Algebra, Answers

Question: Cubes of three Consecutive Integers

Saturday, 2 of February , 2008 at 8:16 pm

Question: Show that the sum of the cubes of any three consecutive integers is divisible by 9.

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Category: Algebra, Questions

Answer: Prove function has even values

Friday, 25 of January , 2008 at 9:48 pm

Question: Prove that the function n^2 -n has even values for all natural numbers (\mathbb{N} = \{0, 1, 2, 3 \ldots\}).

* Updated question for clarity.

Answer: Solving this problem is really trivial. The easiest way (by far) to solve this problem is to use Mathematical induction.

Wikipedia describes Mathematical Induction as:

The simplest and most common form of mathematical induction proves that a statement holds for all natural numbers n and consists of two steps:

  1. The basis: showing that the statement holds when n = 0.
  2. The inductive step: showing that if the statement holds for n = m, then the same statement also holds for n = m + 1.

This is exactly what we need to do to solve the problem. First, we need to prove that f(0) is even. This can easily be done:

f(0) = 0^2 - 0 = 0

Therefore, since 0 is an even number, we have satisfied the first step.

The second step is to show that if f(n) is even than f(n + 1) is also even.

f(n+1) = (n + 1)^2 - (n+1)
= (n^2 + 2n + 1) - (n+1)
= (n^2 -n) + 2n
= f(n) + 2n

If we assume f(n) is even, we know that f(n) + 2n has to be even because even + even = even.

Therefore, the function f(n) = n^2 -n is even for all n \ge 0 because:

  1. f(0) is even.
  2. If f(n) is even, it implies that f(n +1) is even.

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Category: Algebra, Answers