❓ Question

Evaluate the sum of the series:

S = 2 \times 1 \times (\binom{20}{4}) - 3 \times 2 \times (\binom{20}{5}) + 4 \times 3 \times (\binom{20}{6}) - 5 \times 4 \times (\binom{20}{7}) + \dots + 18 \times 17 \times (\binom{20}{20}) .

🖼️ Question Image

(- 1)^{r} (r + 1) r (\binom{20}{r + 3}), r = 1, 2, \dots, 17.

(r + 1) r (\binom{20}{r + 3}) = (r + 3) (r + 2) (r + 1) r \cdot \frac{20!}{(r + 3)! (20 - r - 3)!} .

Better trick: Use identity:

r (r + 1) (\binom{20}{r + 3}) = 20 \times 19 \times (\binom{18}{r + 1}) .

S = \sum_{r = 1}^{17} (- 1)^{r} \cdot 20 \times 19 \times (\binom{18}{r + 1}) .

Factor constants:

S = 380 \sum_{r = 1}^{17} (- 1)^{r} (\binom{18}{r + 1}) .

S = 380 \sum_{k = 2}^{18} (- 1)^{k - 1} (\binom{18}{k}) .

S = - 380 \sum_{k = 2}^{18} (- 1)^{k} (\binom{18}{k}) .

But recall Binomial theorem:

\sum_{k = 0}^{18} (- 1)^{k} (\binom{18}{k}) = (1 - 1)^{18} = 0.

So:

\sum_{k=2}^{18} (-1)^k \binom{18}{k} = -\left[ \binom{18}{0}(-1)^0 + \binom{18}{1}(-1)^1 \right].

= -(1 - 18) = 17.

S = - 380 \times 17 = - 6460.

Hence, the sum of the series is:

- 6460 ​