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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 ?

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

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  • $\begingroup$ Can you please elaborate why took a form as $A.\frac{n(n+1)}{2} +1 = (Bn+C)^2$. $\endgroup$ – jiten Mar 23 '18 at 15:17
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    $\begingroup$ If you can write them in the form of $(Bn+C)^2$, then it is a square number, here $A$ is $8$ and $B, C \in \mathbb{Z}$. $\endgroup$ – Siong Thye Goh Mar 23 '18 at 16:45
  • $\begingroup$ @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. $\endgroup$ – jiten Mar 23 '18 at 17:44
  • $\begingroup$ @SiongThyeGoh Please at least upvote my comment, if no further comment in response is given to make me feel better with my comment. $\endgroup$ – jiten Mar 24 '18 at 0:57
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    $\begingroup$ Alternatively $8T_n$ is even and hence $8T_n+1$ must be odd. Also, you might want to write $8T_n+1$ rather than $8n+1$. $\endgroup$ – Siong Thye Goh Mar 24 '18 at 18:02
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$$\sum_{i=1}^ki=\sum_{i=1}^k\frac{(i+1)^2-i^2-1}{2}=\frac{(k+1)^2-1^2-k}{2}=\frac{k(k+1)}{2}$$

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

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

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

enter image description here

Remark:

Each bigger block contains the smaller block.

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

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