Simple Induction Proof: $\sum_{i=1}^{n} = \frac{i(i+1)}{2}$

Question

I want to prove by induction that $\sum_{i=1}^{n} = \frac{i(i+1)}{2}$

I'm getting stuck as to whether this can be done. My base case of 1 says no.

This is a relatively straightforward induction proof. Why does the base case of $1$ not work? The sum from 1 to 1 is 1 which is equal to $\frac{(1)(2)}{2}$ — Harry Alli, Nov 09 '17 at 13:44
There are two errors here. The equation should be $\sum_{i=1}^n i=\frac{n(n+1)}{2}$. — Rick Decker, Nov 09 '17 at 15:42
I actually messed up with writing the question here and made another post because of it. The summation is suppose to be ∑ni=1 1/i(i+1) — kayin60, Nov 09 '17 at 16:07

score 1 · Answer 1 · answered Nov 09 '17 at 13:44

For n=1, we get:

$\sum_{i=1}^1 i=1=\frac{1\cdot 2}{2}=1\checkmark$

For the inductive step $n\mapsto n+1$:

$\sum_{i=1}^{n+1} i=\sum_{i=1}^n i+(n+1)=\frac{n(n+1)}{2}+(n+1)$ by assumption.

We obtain: $\frac{n(n+1)+2(n+1)}{2}=\frac{(n+1)(n+2)}{2}\checkmark$

1 Answers1