Fibonacci sequence and Pascal triangle

In previous post, we used the result of a tile matching puzzle to prove an identity for Fibonacci sequence. Today, we will continue on and prove another identity. We will show that $${2011 \choose 0} + {2010 \choose 1} + {2009 \choose 2}+ {2008 \choose 3}+ \dots + {1007 \choose 1004}+ {1006 \choose 1005} = F_{2012},$$ $${2012 \choose 0} + {2011 \choose 1} + {2010 \choose 2}+ {2009 \choose 3}+ \dots + {1007 \choose 1005}+ {1006 \choose 1006} = F_{2013}.$$
In general, we have the following identity $$\sum_{v+u=n}{v \choose u} = F_{n+1}.$$ Through this identity, we will see an interesting connection between the Fibonacci sequence and the Pascal triangle.

Fibonacci identities

In previous post, we have learnt about Fibonacci sequence and a tile matching puzzle. Today, we will use the result of the tile matching puzzle to prove the following identity on Fibonacci sequence $$F_{n+m+2} = F_{n+1} F_{m+1} + F_{n+1} F_m + F_n F_{m+1}.$$

Like an art painting, a mathematical problem reflects different beauties if we view it at different angles. Take, for example, the Fibonacci identity that we are considering, we can prove this identity by different methods. One method is to use the formula $$F_n = \frac{1}{\sqrt{5}} \left[ \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right]$$ to calculate the two sides of the identity and make them equal by straightforward algebraic manipulations. Another method is to make use of the tile matching puzzle that we learned in the previous post. I hope that you will find this "combinatorial" method more interesting.

Fibonacci numbers and a tile matching puzzle

Today we will learn about Fibonacci sequence and a tile matching puzzle. We will soon see that the tile matching puzzle seems to be unrelated to the Fibonacci sequence, but it turns out that the solution to the puzzle is, in fact, the Fibonacci sequence!

Let us first introduce the Fibonacci sequence. The Fibonacci sequence $\{F_n\}$ is determined by the following rule: $$F_0 = 0, ~F_1 = 1, ~F_{n+1} = F_n + F_{n-1}.$$

Thus, $$F_0 = 0, ~F_1 = 1, ~F_2 = 1, ~F_3 = 2, ~F_4 = 3, ~F_5 = 5, ~F_6 = 8, ~F_7 = 13, ~F_8 = 21, \dots$$

1 = 2012 = 2013

In previous posts, we learned about mathematical induction and we used induction to solve some problems. We can see that mathematical induction is a useful technique in problem solving. Today, we will consider two induction proofs that lead to a wrong result that $$1 = 2012 = 2013$$ Let us know if you can identify the wrong steps in the proofs.

Binomial identity

As we are learning about mathematical induction, in this post, we are going to use induction to prove the following:

• formula for the Pascal's triangle $$p_{n,k} = {n \choose k} = \frac{n!}{k! (n-k)!}$$
• the binomial identity $$(x+y)^n = x^n + {n \choose 1} x^{n-1} y + {n \choose 2} x^{n-2} y^2 + \dots + {n \choose {n-2}} x^{2} y^{n-2} + {n \choose {n-1}} x y^{n-1} + y^n$$

Mathematical induction III

Today, we will solve some more problems using mathematical induction.

Problem 7. Observe that $$\cos 2 \alpha = 2 \cos^2 \alpha - 1$$

Prove that we can write $\cos n\alpha$ as a polynomial of $\cos \alpha$.

Mathematical induction II

Today we will use induction to solve some more problems. 

Problem 4. Prove that $$1 \times 2 \times 3 + 2 \times 3 \times 4 + \dots + n (n+1)(n+2) = \frac{1}{4} n(n+1)(n+2)(n+3).$$

Mathematical induction

Today, we will learn about mathematical induction. We usually use induction to prove a certain statement to be correct for all natural numbers.

Let us use $P(n)$ to denote a statement that involves a natural number $n$. To prove that $P(n)$ is correct for all natural number $n$, an induction proof will have the following steps

Step 1: is called the initial step. We will prove that the statement $P(n)$ is correct for the case $n=0$.

Step 2: is called the induction step. This is the most important step. In this step,

• we assume that for any $0 \leq n \leq k$, the statement $P(n)$ is correct;
• with this assumption, we will prove that the statement $P(n)$ is also correct for the case $n=k+1$.

With these two steps, by the mathematical induction principle, we conclude that the statement $P(n)$ must be correct for all natural number $n$.