Square of Triangular Numbers as Sum of Triangular Numbers

Theorem

${T_n}^2 = T_n + T_{n - 1} T_{n + 1}$

where $T_n$ denotes the $n$th triangular number.

$\ds T_n + T_{n - 1} T_{n + 1}$

$=$

$\ds \frac {n \paren {n + 1} } 2 + \paren {\frac {\paren {n - 1} n} 2 \frac {\paren {n + 1} \paren {n + 2} } 2}$

$\ds $

$=$

$\ds \frac {2 n^2 + 2 n + \paren {n^2 - n} \paren {n^2 + 3 n + 2} } 4$

$\ds $

$=$

$\ds \frac {2 n^2 + 2 n + n^4 - n^3 + 3 n^3 - 3 n^2 + 2 n^2 - 2 n} 4$

$\ds $

$=$

$\ds \frac {n^4 + 2 n^3 + n^2} 4$

$\ds $

$=$

$\ds \paren {\frac {n \paren {n + 1} } 2}^2$

$\ds $

$=$

$\ds {T_n}^2$

$\blacksquare$