Q28 / 75 · JEE Main 2025 (7 Apr Shift 2) · Mathematics
The sum of the series
$2 \times 1 \times{ }^{20} \mathrm{C}_4-3 \times 2 \times{ }^{20} \mathrm{C}_5+4 \times 3 \times{ }^{20} \mathrm{C}_6-5 \times 4$ $\times { }^{20} \mathrm{C}_7+\ldots+18 \times 17 \times{ }^{20} \mathrm{C}_{20}$, is equal to
Correct answer: 34
Solution
$\begin{aligned}
& (1-\mathrm{x})^{20}={ }^{20} \mathrm{C}_0-{ }^{20} \mathrm{C}_1 \mathrm{x}+{ }^{20} \mathrm{C}_2 \mathrm{x}^2 \ldots . .+{ }^{20} \mathrm{C}_{20} \mathrm{x}^{20} \\
& \frac{(1-\mathrm{x})^{20}}{\mathrm{x}^2}=\frac{{ }^{20} \mathrm{C}_0}{\mathrm{x}^2}-\frac{{ }^{20} \mathrm{C}_1}{\mathrm{x}}+{ }^{20} \mathrm{C}_2-{ }^{20} \mathrm{C}_3 \mathrm{x}+{ }^{20} \mathrm{C}_4 \mathrm{x}^2 \ldots
\end{aligned}$
Diff twice and put $\mathrm{x}=1$
$\begin{aligned}
& =6-{ }^{20} \mathrm{C}_1(2)+\mathrm{A} \\
& \mathrm{~A}=40-6=34
\end{aligned}$
Hence, the correct answer is 34.