The required is to prove that $\sum_{i=1}^n 1 = n$
So here's my attempt
- Prove that it works for n = 1
$$\sum_{i=1}^1 1 = 1$$ $$ 1 = 1$$
- Assume it works for k
$$\sum_{i=1}^k 1 = k$$
- Show that it works for k + 1
$$(\sum_{i=1}^k 1)= k$$ (adding 1 to both sides) $$(\sum_{i=1}^k 1) + 1 = k + 1$$ Since ($\sum_{i=1}^k 1$) is equal to (k) from step 2 we can do the substitution (This is the step I'm not sure of because It seems like I'm using the rule I'm trying to prove) $$k + 1 =k + 1$$
Step 3 in your argument is not quite right. If you are trying to prove a statement $S(n)$, then you do as you did for the base case, then you fix some $k\geq1$, and show that $S(k)\to S(k+1)$, where $S(k)$ is your inductive hypothesis. To that end, your argument should not be $$ \sum_{i=1}^k1=k\Leftrightarrow\sum_{i=1}^k1+1=k+1,\tag{1} $$ as it is currently (where you appear to be working backwards almost), but something more like $$ \sum_{i=1}^{k+1}1=\underbrace{\sum_{i=1}^k1+1}_{\text{by definition}}=\underbrace{k+1}_{\text{by inductive hyp.}},\tag{2} $$ where you explicitly use the inductive hypothesis to show that $S(k)\to S(k+1)$, as in $(2)$.