A general formula for $\sum (k-1)(k-2)(k-3)$?

sequences-and-series algebra-precalculus

What is a "simpler" formula for

$$\sum_{3}^{n} \frac{(k-1)(k-2)(k-3)}{6}$$


Solution 1:

Hint:

$$k(k-1)(k-2)(k-3) - (k-1)(k-2)(k-3)(k-4) = 4(k-1)(k-2)(k-3)$$

Solution 2:

Show that ${k+3 \choose 3}$ is the number of solutions to $x_1 + ... + x_4 = k$ in non-negative integers. Then $\sum_{k=0}^{n-4} {k+3 \choose 3}$ is the number of solutions to $x_1 + ... + x_5 = n-4$ in non-negative integers, which is...?

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