$\begin{aligned}
& (1-\mathrm{x})^{20}={ }^{20} \mathrm{C}_0-{ }^{20} \mathrm{C}_1 \mathrm{x}+{ }^{20} \mathrm{C}2 \mathrm{x}^2 \ldots . .+{ }^{20} \mathrm{C}{20} \mathrm{x}^{20} \
& \frac{(1-\mathrm{x})^{20}}{\mathrm{x}^2}=\frac{{ }^{20} \mathrm{C}_0}{\mathrm{x}^2}-\frac{{ }^{20} \mathrm{C}_1}{\mathrm{x}}+{ }^{20} \mathrm{C}_2-{ }^{20} \mathrm{C}_3 \mathrm{x}+{ }^{20} \mathrm{C}_4 \mathrm{x}^2 \ldots
\end{aligned}Difftwiceandput\mathrm{x}=1\begin{aligned}
& =6-{ }^{20} \mathrm{C}_1(2)+\mathrm{A} \
& \mathrm{~A}=40-6=34
\end{aligned}$
The sum of the series 2 × 1 × ^20 C_4-3 × 2 × ^20 C_5+4 × 3 × ^20 C_6-5 × 4 × ^20 C_7+…+18 × 17 × ^20 C_20, is equal to
JEE Main 2025 — Mathematics Algebra
The sum of the series 2×1×20C4−3×2×20C5+4×3×20C6−5×4 ×20C7+…+18×17×20C20, is equal to
Held on 7 Apr 2025 · Verified 6 Jul 2026.
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