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I saw some video where it needs to prove $1+2+\dots+n=\frac{n(n+1)}{2}$ inductively. So it has to be true if $k=1$ and $k+1$ are true.

So, for $k=1$:

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

it is valid.

For $k+1$ here is the proof he does:

$$ \begin{align} 1+2+\dots+k+k+1&=\frac{(k+1)(k+2)}{2} &(1)\\ \frac{k(k+1)}{2}+k+1&=\frac{(k+1)(k+2)}{2} &(2)\\ &=\frac{k^2+2k+k+2}{2}&(3)\\ &=\frac{k^2+3k+2}{2}&(4)\\ &=\frac{(k+1)(k+2)}{2}&\text{factoring (4)} \end{align}$$

Therefore this formula is valid for $k+1$.

But is this true? I think not. He is just undoing what he have just done. To prove it I think I need to do this:

$$ \begin{align} 1+2+\dots+k+k+1&=\frac{(k+1)(k+2)}{2}\\ \frac{k(k+1)}{2}+k+1&=\frac{(k+1)(k+2)}{2}\\ \frac{k(k+1)+2(k+1)}{2}&=\frac{(k+1)(k+2)}{2}\\ \frac{(k+1)(k+2)}{2}&=\frac{(k+1)(k+2)}{2}\\ (k+1)(k+2)&=(k+1)(k+2)\\ k+1&=k+1\\ k&=k\\ 0&=0\text{ or }1=1 \end{align}$$

Therefore, $k+1$ is valid in this formula.

Is this right or am I making a mistake somewhere?

Garmekain
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4 Answers4

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The second way is correct but I would be more carefull and use like this:

$$1+2+...+k+(k+1)=\frac{k(k+1)}{2}+k+1=\frac{k^2+3k+2}{2}=\frac{(k+1)(k+2)}{2}$$

P.s: Not use the equality in the first place. The problem using the equality is that you need guarantee equivalence in every step what is not so easy in many problems.

Arnaldo
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Proving for sum =k+1

RhS= $\dfrac {(k)(k+1)}{2} + k + 1 $

Which gives

=$\dfrac {(k+1)(k+2)}{2}$

= $\dfrac {([k+1])([k+1] +1)}{2}$

Which means it hold for n and also n+1 !

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This proof is at least unlucky written down.

Normally, we start with the claim for $n$ (Here $1+2+\cdots n=\frac{n(n+1)}{2}$) and then proof the claim for $n+1$.

The correct way is to start with $1+2+\cdots n+n+1=\frac{n(n+1)}{2}+n+1$ (we assume that the formula is correct for $n$) and show that this is equal to $\frac{(n+1)(n+2)}{2}$ by simply adding the two fractions.

It is unusual to start with the aim of the induction step and then going back to show that it is valid.

Peter
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You don't to begin using equality!! You must to arrive in this... $1+2+⋯+k+k+1= $(here you use the induction hipoteses) Than...We start correct:

$$1+2+⋯+k+k+1=**induction hipoteses(ih)**\frac{k(k+1)}{2}**finish(ih)**+**hereyourepeatlikeabove**k+1$$

I hope I've helped.