Triangular series perfect square formula 8n+1 derivation In triangular series 
$$1 $$
$$1+2 = 3$$
$$1+2+3 = 6$$
$$1+2+3+4 =10$$
$$\ldots$$
Triangular number in  8n+1 always form perfect square .
Like $8\cdot 1+1 = 9 , 8\cdot 3+1 = 25$ .
How this formula is derived ?

$$\sum_{i=1}^ki=\sum_{i=1}^k\frac{(i+1)^2-i^2-1}{2}=\frac{(k+1)^2-1^2-k}{2}=\frac{k(k+1)}{2}$$
The $k$th triangular number is $\frac{k(k+1)}{2}$.
$$8\left[\frac{k(k+1)}{2}\right] +1=4k^2+4k+1=(2k+1)^2$$
