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
 A: 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
A: The most transparent and general way to deal with these questions is with generating functions. The humble geometric series identity says:
$$1+x+\cdots+x^{n-1} = \frac{1-x^n}{1-x}$$
Note that $\frac{d^2}{dx^2} x^i = i(i-1) x^{i-2}$. We can get a sum of triangular number coefficients by differentiating the above twice, dividing by $2$, and setting $x=1$. You want squares of triangular numbers though. For that, just notice
$$\left(x^2 \frac{d^2}{dx^2} x^2 \frac{d^2}{dx^2}\right) x^i = x^2\frac{d^2}{dx^2} x^2 i(i-1)x^{i-2} = i^2 (i-1)^2 x^i. $$
Thus
$$\begin{align*}
\sum_{i=1}^n \left(\frac{i(i+1)}{2}\right)^2
&= \sum_{i=0}^{n+1} \left(\frac{i(i-1)}{2}\right)^2 \\
&= \sum_{i=0}^{n+1} \left.\left(\frac{i(i-1)}{2}\right)^2 x^i\right|_{x=1} \\
&= \frac{1}{4} \sum_{i=0}^{n+1} \left.\left(x^2 \frac{d^2}{dx^2} x^2 \frac{d^2}{dx^2} x^i\right)\right|_{x=1} \\
&= \frac{1}{4} \left.\left(x^2 \frac{d^2}{dx^2} x^2 \frac{d^2}{dx^2} \sum_{i=0}^{n+1} x^i\right)\right|_{x=1} \\
&= \frac{1}{4} \lim_{x \to 1} \left(x^2 \frac{d^2}{dx^2} x^2 \frac{d^2}{dx^2} \frac{1-x^{n+2}}{1-x}\right) \\
&= \cdots \\
&= \frac{n (1 + n) (2 + n) (1 + 3 n (2 + n))}{60}
\end{align*}$$
Here $\cdots$ hides the mess. A few remarks are in order:

*

*No special identities are required--no random Faulhaber formula, sum of squares, sum of cubes, etc. You just need to churn through routine calculus.

*The step right before the messy calculation tells you all you really  to know: it's very unlikely the final answer will have a nice form, since that messy nest of quotient rule applications won't.

*If you really desperately do want to know the explicit final answer, a computer can tell it to you with no margin for error or additional effort: in Mathematica, 

*If you decide you want variations on this, e.g. cubes of triangular numbers rather than squares, it's often trivial to modify the argument and have a computer tell you the new answer.

