<s>[INST] Find the value of this combinatorial sum. $\sum_{k=4}^{100}\binom{k-1}{3}$ How to compute this sum without laboring?
$$\sum_{k=4}^{100}\dbinom{k-1}{3}.$$
Is it possible to reduce this to a single combinatorial term?
 [/INST] $$\sum_{k=4}^{100}\dbinom{k-1}{3}  = \sum_{k=4}^{100} \frac{(k-1)(k-2)(k-3)}{3!} = \sum_{k=4}^{100}\frac{k^3  - 6k^2 + 11k - 6}{6} \\ = \frac{1}{6} \sum_{k=4}^{100} \left(k^3 - 6k^2 + 11 k - 6\right) = \frac{1}{6} \sum_{k=1}^{100} \left(k^3 - 6k^2 + 11 k - 6\right)-  \frac 1 6 \sum_{k=1}^{3} \left(k^3 - 6k^2 + 11 k - 6\right)$$
then perhaps look into sum of cubes and squares formula??
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