Centered triangular number

Statement

The $n$-th centered triangular number is $c_n=3T_n+1$, with $T_n=\dfrac{n(n+1)}{2}$. Compute $c_6$ and note it is a perfect square.

Solution

$T_6=\frac{6\cdot 7}{2}=21,\qquad c_6=3\cdot 21+1=\boxed{64}=8^2.$