Can someone look at my proof. I am supposed to prove by induction. The question is to prove the following:
$$\sum _{i=0}^{n}{i} =\frac { n\left( n+1 \right) }{ 2 } .$$
If $n=1$ Then
$$1=\sum_{i=1}^1{i}=\frac{1(1+1)}{2}=1$$
Now assume $n=k$. Thus
$$\sum_{i=1}^k{i}=\frac{k(k+1)}{2}=1$$
If now $n=k+1$
$$\sum_{i=1}^{k+1}{i}=\sum_{i=1}^k{i}+(k+1)= \frac{(k+1)(k+2)}{2}$$
Is there anything more that I have to do, or is this it?
You've got it. Some might pick a nit saying that you forgot to say that this shows that the proposition is true for $n=k+1$, thus establishing induction. But I would give full credit for your proof.
You assume that the statement is true for $n=k$. In other words, assuming $$ \sum_{i=1}^{k}{i}=\frac{k(k+1)}{2} $$ is true. I don't know why you set it to $1$ when $n=k$. I assume that it's a typo. Next, we'll prove that the statement is true for $n=k+1$ $$ \sum_{i=1}^{k+1}{i}= \sum_{i=1}^{k}{i}+(k+1)=\frac{k(k+1)}{2}+(k+1)=\frac{(k+1)(k+2)}{2} $$ where you use the assumption that $\sum_{i=1}^{k}{i}=\frac{k(k+1)}{2}$. Then you're done, because this is true for all $n \in \mathbb{N}$.
I guess you could add the middle equality to the last line,so it's clear that you used the assumption $$\sum_{i=1}^{k+1}i=\sum_{i=1}^{k}i+(k+1)=\frac{k(k+1)}{2}+(k+1)=\frac{(k+1)(k+2)}{2}$$ Other than that everything is ok.
