Proof that $\sum_{i=1}^n{1} = n$ for all $n \in \Bbb Z^+$

Question

It seems obvious that $$\forall n \in \Bbb Z^+, \sum_{i=1}^n{1} = n $$

However, I'm having trouble coming up with a formal proof for this.

Given a concrete number like $4$, we can say that $$\sum_{i=1}^4{1} = 1+1+1+1 = 2+1+1=3+1=4$$ by the definitions of summation, 2, 3, and 4. But how can we formally show this for an arbitrary positive integer?

I wonder if we can use the definitions of multiplication and the multiplicative identity, but I'm not sure.

Answer 1

Hint: Use induction.

To get you started, suppose we have the statement $$ S(n) : \sum_{i=1}^n 1 = n $$ that you are trying to prove. Fix some $k\geq 1$ and assume $$ S(k) : \sum_{i=1}^k 1 = k $$ is true. Then we need to show that $$ S(k+1) : \sum_{i=1}^{k+1} 1 = k+1 $$ follows. Beginning with the left side of $S(k+1)$, \begin{align} \sum_{i=1}^{k+1} 1 &= \sum_{i=1}^k 1 + 1\tag{using definition of $\Sigma$}\\[1em] &= k + 1\tag{by $S(k)$}, \end{align} we reach the right side of $S(k+1)$.

The result, then, follows by mathematical induction.