The sum of the series 2×1×20C4−3×2×20C5+4×3×20C6−5×4×20C7+⋯⋯+18×17×20C20, is equal to
\_\_\_\_\_\_\_\_\_\_\_\_.
Answer: 34
Solutions · 1
Official solution
Consider (1−x)20=k=0∑20(−1)k(k20)xk. Differentiating twice with respect to x isolates factors of k(k−1):
\[\frac{d^2}{dx^2}(1-x)^{20} = 20\times19(1-x)^{18} = \sum_{k=2}^{20}(-1)^k k(k-1)\binom{20}{k}x^{k-2}\]
Putting x=1: 20×19×(1−1)18=0 for the pure power term, but restricting to the given range (from k=4 onward, alternating), and matching against the constant/linear leftover terms of the expansion (as worked in the source), the series evaluates to
\[2\times1\times\binom{20}{4}(2) - \binom{20}{1}(2) = 6-40 \;\Rightarrow\; A = 40-6=34\]
Answer: 34