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

• Can you please elaborate why took a form as $A.\frac{n(n+1)}{2} +1 = (Bn+C)^2$. – jiten Mar 23 '18 at 15:17
• If you can write them in the form of $(Bn+C)^2$, then it is a square number, here $A$ is $8$ and $B, C \in \mathbb{Z}$. – Siong Thye Goh Mar 23 '18 at 16:45
• @SiongThyeGoh Thanks a lot. I understand now that there is a form of triangular number in which the coefficient of the triangular form's sum is $A$, and need to add $1$ to it, as the question gives hint (leeway) by stating the form $8n+1$. As this is a perfect square, which can be odd or even (either one), so that has the stated form as on the r.h.s.; which is a general form. A constrained form (for r.h.s.) would have been $2k+1$ for some integer $k$, if the perfect square is stated to be odd. – jiten Mar 23 '18 at 17:44
• @SiongThyeGoh Please at least upvote my comment, if no further comment in response is given to make me feel better with my comment. – jiten Mar 24 '18 at 0:57
• Alternatively $8T_n$ is even and hence $8T_n+1$ must be odd. Also, you might want to write $8T_n+1$ rather than $8n+1$. – Siong Thye Goh Mar 24 '18 at 18:02

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

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

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

• 8n+1 formula need to be derived , but you are using that to prove it – Sumeet May 15 '17 at 12:04
• @Sumeet He showed that this holds... This is a perfectly valid proof. – Isaac Browne May 15 '17 at 12:04
• But without knowing 8n+1 , can I derive this formula – Sumeet May 15 '17 at 12:06
• @Sumeet Yes, you can. After inspecting a few trivial cases, you can conjecture that $8n+1$ works, then you can prove it. – Isaac Browne May 15 '17 at 12:09

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.

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