Step-by-step explanation:
We will prove by mathematical induction that, for every natural n,
[tex]\sum^{n}_{i=1}i=\frac{n(n+1)}{2}[/tex]
We will prove our base case to be true:
Base case:
For n=1,
[tex]\sum^{n}_{i=1}i=\sum^{1}_{i=1}i=1=\frac{1(1+1)}{2}=\frac{n(n+1)}{2}.[/tex]
Inductive hypothesis:
Given a natural n,
[tex]\sum^{n}_{i=1}i=\frac{n(n+1)}{2}[/tex]
Now, we will assume the inductive hypothesis and then use this assumption, involving n, to prove the statement for n + 1.
Inductive step:
Observe that,
[tex]\sum^{n+1}_{i=1}i=\sum^{n}_{i=1}i+(n+1)=\frac{n(n+1)}{2}+(n+1)=\\\\=\frac{n(n+1)+2(n+1)}{2}=\frac{(n+2)(n+1)}{2}.[/tex]
With this we have proved our statement to be true for n+1.
In conclusion, for every natural n,
[tex]\sum^{n}_{i=1}i=\frac{n(n+1)}{2}[/tex].
