Sum of squares of consecutive triangular numbers

Question

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

Answer 1

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

Answer 2

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:

1. No special identities are required--no random Faulhaber formula, sum of squares, sum of cubes, etc. You just need to churn through routine calculus.
2. 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.
3. 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,
4. 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.