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Proof for the Partial Sum Formula of i

Looking for a proof for this formula $ sum_{i=1}^n i = frac{n(n+1)}{2} $
Looking for a proof using induction as it seems to be simpler to me, rather than a combinatorical one.


Suppose you have an induction on $n$. Then assume the following :

$n=1$ is true
$n+1$ is true
The difference between $n$ and $n+1$ is $1$

What does this imply? For $n=1$, the LHS is 1, and the RHS is $dfrac{1(1+1)}{2} = dfrac{1cdot 2}{2} = 1$.
For $n+1$ the LHS is $dfrac{1(1+1)}{2}$; while the RHS is $dfrac{(n+1)(n+2)}{2} = dfrac{(1+1)cdot(n+1)}{2}$.
So what does this imply? It implies that $dfrac{(1+1)(n+1)}{2} = dfrac{(1+1)cdot(n+1)}{2}$.
But $dfrac{(1+1)cdot(n+1)}{2} – (n+1) = dfrac{1}{2}$
And that’s enough.



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