How is derived the inductive step in mathematical induction?

Question

I am quite familiar with the algorithm of mathematical induction but I can't rationalize the inductive step very well. Suppose I have the classical example: $$0 + 1 +2 + \ldots + n = \frac{n(n+1)}{2}$$ In the inductive step I have to show that this hold for $k+1$ so I don't understand why I don't find written: $$0 + 1 +2 + \ldots + (k+1) = \frac{(k+1)((k+1)+1)}{2}$$ instead of: $$0 + 1 +2 + \ldots + k + (k+1) = \frac{(k+1)((k+1)+1)}{2}$$

Why the $k$ in the left-hand side of the equation in not substitute with $k+1$ as in the right-hand side of the equation. How can I rationalize the inductive step of mathematical induction? Can any one prove me that the first equation is also right and show me how to deal with the proof?

Answer 1

Can anyone prove me that the first equation is also right and show me how to deal with the proof?

Apparently you don't understand that the first and second equations are exactly the same,

$$0+1+\cdots+5$$

and $$0+1+\cdots+4+5$$

are the same things.

Answer 2

i don't know what exactly you mean. but the desired claim is that for every $n\in\mathbb{N}$ such $n=k+1\in\mathbb{N}$, the equality holds .

Now, in Inductive step, you suppose the proposition is true for $k$ and then you must prove the proposition for $k+1$. It means you must prove that:

$$1+\cdots+(k+1)=\frac{(k+1)((k+1)+1)}{2}$$

It's now easy to check above equality as :

$$1+\cdots+k+(k+1)=\frac{k(k+1)}{2}+k+1$$

Answer 3

Using induction you have $1 + 2 + \cdots + k = \frac{k(k + 1)}{2}$. Butthen you can write $1 + 2 + \cdots + k + (k + 1) = \frac{k(k + 1)}{2} + (k + 1)$. Now $$\frac{k(k + 1)}{2} + (k + 1) = (k + 1)(\frac{k}{2} + 1) = (k + 1)\frac{k + 2}{2} = \frac{1}{2}(k + 1(k + 2).$$ This is what you need to show.