2024/030) Prove that for triangular numbers $t_{3k}+t_{4k+1}=t_{5k+1}$

We have by definition  triangular number $t_k = \frac{k(k+1)}{2}$

So   $t_{3k}+t_{4k+1}$

$= \frac{(3k(3k+1)}{2} + \frac{(4k+1)(4k+2)}{2}$

$= \frac{(9k^2+3k}{2} + \frac{16k^2 + 12 k + 2}{2}$

$= \frac{25k^2+15k + 2}{2}$

$= \frac{(5k+1)(5k + 2)}{2}= t_{5k+1}$