1+2+...+n+(n-1)+...+1 = n^2

$$1+2+...+n+(n-1)+...+1 = n^2$$, $$\forall n\in N$$

Direct Proof
$$1+2+\dots+n+(n-1)+\dots+1$$

$$=1+2+\dots+(n-1)+(n-1)+(n-2)+\dots+1+n$$

$$=2(1+2+\dots+(n-1))+n$$

$$=2(\frac{(n-1)n}{2})+n$$ by 1+2+...+n = n(n+1)/2.

$$=(n-1)n+n$$

$$=n^2-n+n$$

$$=n^2$$ as desired.

Q.E.D.

Proof by Induction
Base case: n=2

$$1+2+1 = 4$$, likewise $$n^2=2^2=4$$. So shown for base case.

Induction Hypothesis:

$$1+2+\dots+k+(k-1)+\dots+1=k^2$$

Show true for $$n=k+1$$:

Want to show that, $$1+2+\dots+(k+1)+k+(k-1)+\dots+1=(k+1)^2$$

So, $$1+2+\dots+(k+1)+k+(k-1)+\dots+1=(1+2+\dots+k+(k-1)+\dots+1)+k+(k+1)$$, since $$1+2+\dots+k+(k-1)+\dots+1=k^2$$

Then, $$k^2+k+(k+1)=k^2+2k+1=(k+1)^2$$

Q.E.D.