Triangular series perfect square formula 8n+1 derivation

Solution 1:

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 $$8T_n+1=(2n+1)^2$$

Solution 2:

The other answers have provided algebraic reason why

$$8T_n+1$$ is a square.

That is because we can simplify it into $$8T_n+1=(2n+1)^2$$

Geometrically, it means we can put a special block in the middle of the square. The special block has $8$ neighbors, we can put a vertex of the triangle at each such block and arrange them such that the each side of the square consists of two components of sides of a triangle and a corner of another triangle.

Remark:

Each bigger block contains the smaller block.

$$8T_n+1=(2n-1)^2+8n$$

Solution 3:

$$\sum_{i=1}^ki=\sum_{i=1}^k\frac{(i+1)^2-i^2-1}{2}=\frac{(k+1)^2-1^2-k}{2}=\frac{k(k+1)}{2}$$

The $k$th triangular number is $\frac{k(k+1)}{2}$.

$$8\left[\frac{k(k+1)}{2}\right] +1=4k^2+4k+1=(2k+1)^2$$