JEE MAIN - Mathematics (2025 - 7th April Evening Shift - No. 21)
The sum of the series $2 \times 1 \times{ }^{20} \mathrm{C}_4-3 \times 2 \times{ }^{20} \mathrm{C}_5+4 \times 3 \times{ }^{20} \mathrm{C}_6-5 \times 4 \times{ }^{20} \mathrm{C}_7+\cdots \cdots+18 \times 17 \times{ }^{20} \mathrm{C}_{20}$, is equal to
____________.
Answer
34
Explanation
$$\begin{aligned}
&\begin{aligned}
& (1-\mathrm{x})^{20}={ }^{20} \mathrm{C}_0-{ }^{20} \mathrm{C}_1 \mathrm{x}+{ }^{20} \mathrm{C}_2 \mathrm{x}^2 \ldots . .+{ }^{20} \mathrm{C}_{20} \mathrm{x}^{20} \\
& \frac{(1-\mathrm{x})^{20}}{\mathrm{x}^2}=\frac{{ }^{20} \mathrm{C}_0}{\mathrm{x}^2}-\frac{{ }^{20} \mathrm{C}_1}{\mathrm{x}}+{ }^{20} \mathrm{C}_2-{ }^{20} \mathrm{C}_3 \mathrm{x}+{ }^{20} \mathrm{C}_4 \mathrm{x}^2 \ldots .
\end{aligned}\\
&\text { Diff twice and put } \mathrm{x}=1\\
&\begin{aligned}
& =6-{ }^{20} \mathrm{C}_1(2)+\mathrm{A} \\
& \mathrm{~A}=40-6=34
\end{aligned}
\end{aligned}$$
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