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Can someone explain to me this :

Why we use both, "n" and "n+1" in the third stage if math induction (where we check if statement holds for "n+1". I'll give an example

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

When I get to the third step, I should write : $1+2+3+...+n+(n+1)=$ doesn't matter My question is, why I can't write $1+2+3+...+(n+1)$? If I must write $n$ in front of $n+1$, shouldn't I write $(n-1)$ where it goes to $n$ ? I hope that you will understand my question.

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That's usually done to remark that you have $\;n\;$ there and up to there you can assume so and so. – DonAntonio Sep 27 '13 at 17:14
@DonAntonio, but if I leave out n, and use 1+2+3...+n+1, I can find 1+2+-.--+n and change it with something, but that will get me wrong result. – BTestQ Sep 27 '13 at 17:17
That's because you're not taking into account that $\;n\;$ is there before $\;n+1\;$ , even if you don't write it explicitly! – DonAntonio Sep 27 '13 at 17:18
@DonAntonio, thank you very much, now I understand. – BTestQ Sep 27 '13 at 17:28
up vote 5 down vote accepted

For example: prove that


So you prove for $\;1\;$ and you assume truthness for $\;n\;$, i.e. you assume the above is true. Now you want to prove for $\;n+1\;$ , so you want to prove


But you write the left side as

$$\underbrace{1+2+...+n}_{\text{you assume this is known!}}+(n+1)\stackrel{\text{inductive hypothesis}}=\frac{n(n+1)}2+(n+1)$$

Now you do the usual algebra on the right side and reach the wanted result...

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@BTestQ: And just to be clear, note that the $n$ term can be left out or show up in the sum, but it makes the logical process more clear if the term that is used is actually visually present in the demonstration. It is not the case that the $n$ term ceases to be part of the sum, but rather that the writer has chosen the shortcut of skipping all the way to the last term. – abiessu Sep 27 '13 at 17:19

Both $1+\dots+(n+1)$ and $1+\dots+n+(n+1)$ mean the same thing, but sometimes the second one is clearer. For example suppose we wrote $$\begin{align*} &1+\dots+n = A\\ \Rightarrow &1+\dots+(n+1) = A+(n+1) \end{align*}$$ This might be a bit confusing. It looks like the "$n$" has turned into an "$n+1$". So it's better to write $$\begin{align*} &1+\dots+n = A\\ \Rightarrow &1+\dots+n+(n+1) = A+(n+1) \end{align*}$$

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