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The required is to prove that $\sum_{i=1}^n 1 = n$

So here's my attempt

  1. Prove that it works for n = 1

$$\sum_{i=1}^1 1 = 1$$ $$ 1 = 1$$

  1. Assume it works for k

$$\sum_{i=1}^k 1 = k$$

  1. 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) $$k + 1 =k + 1$$

ArandomUserNameEG
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    Add line showing $\sum_{i=1}^{k+1} 1 = k + 1$ and you're done. The end of your inductive step should look like your hypothesis / assumption but with +1 – AJJ Aug 08 '16 at 16:50
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    Step 3: $(\sum_{i=1}^{k} 1) + 1 = k + 1$, Step 4: $\sum_{i=1}^{k+1} 1 = k + 1$. – AJJ Aug 08 '16 at 16:52
  • Oh ok, now I understand, I found my error now, thanks, please post it as an answer so I can accept it and close the question – ArandomUserNameEG Aug 08 '16 at 16:53
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    The same kind of question was asked a while back, and you may benefit from looking it over. Since your question certainly falls under the proof-verification tag, I do not see any reason to mark it as a duplicate. – Daniel W. Farlow Aug 08 '16 at 17:36

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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 $(2)$.

Daniel W. Farlow
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