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 ?

Let$$\begin{align}
A\cdot \overbrace{\frac {n(n+1)}2}^{T_n}+1&=(Bn+C)^2\\
\frac A2 n^2+\frac A2n+1&=B^2n^2+2BCn+C^2\\
\end{align}$$
Equating coefficients gives $C=1, A=4B, A=2B^2$, solving for which gives $B=2, A=8$.   
Hence 
$$8T_n+1=(2n+1)^2$$
