Sum of squares of consecutive triangular numbers Find a general formula for:
$t^2_1 + t^2_1 ... t^2_n$.
I tried evaluating:
$1/4[1(2)^2+2^2(3)^2 ... (n)(n+1)]$ then I used sum of cubes to evaluate this.
$(1^3) +(1^3 +2^3) ... (1^3 + 2^3 ... + n^3)$
$n(1^3) +  (n-1)(2^3) ... + n^3(1)$
I also tried grouping terms like:
$n(1)(1^2+n^2) +(n-1)(2)((n-1)^2+2^2)) ...$
but I didn't see anything else

Hint:
Using Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction
$$\sum_{r=1}^n(1^3+2^3+\cdots+r^3)$$
$$=\dfrac14\cdot\left(\sum_{r=1}^nr^4+2\sum_{r=1}^nr^3+\sum_{r=1}^nr^2\right)$$
Use
How to get to the formula for the sum of squares of first n numbers?
and
Sum of fourth powers in terms of sum of squares
