Find the value of this combinatorial sum. $\sum_{k=4}^{100}\binom{k-1}{3}$

Question

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?

Answer 1

In the same spirit as Santosh Linkha $$\sum_{k=4}^{m}\dbinom{k-1}{3} =\frac{1}{24} m ~(m-1)~(m-2)~ (m-3) $$

Answer 2

Hint: you can prove by induction that

$$\sum_{k=p}^{n} {k \choose p}= {n+1 \choose p+1}$$

Proof on an example, and illustration here.

Answer 3

$$\sum_{k=4}^{100}\dbinom{k-1}{3} = \sum_{k=4}^{100} \frac{(k-1)(k-2)(k-3)}{3!} = \sum_{k=4}^{100}\frac{k^3 - 6k^2 + 11k - 6}{6} \\ = \frac{1}{6} \sum_{k=4}^{100} \left(k^3 - 6k^2 + 11 k - 6\right) = \frac{1}{6} \sum_{k=1}^{100} \left(k^3 - 6k^2 + 11 k - 6\right)- \frac 1 6 \sum_{k=1}^{3} \left(k^3 - 6k^2 + 11 k - 6\right)$$

then perhaps look into sum of cubes and squares formula??

Answer 4

By looking at the factors of the sum, I was able to get this: $$\dbinom{100}{4}$$

This answer suggests a proof method.