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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 $n$.