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

Correct answer: 0 aligned & (1-x)^ 20 = ^ 20 C_0- ^ 20 C_1 x+ ^ 20 C_2 x^2 . .+ ^ 20 C_ 20 x^ 20 & (1-x)^ 20 x^2 = ^ 20 C_0 x^2 - ^ 20 C_1 x + ^ 20 C_2- ^ 20 C_3 x+ ^ 20 C_4 x^2 aligned Diff twice and put x=1 aligned & =6- ^ 20 C_1(2)+A & ~A=40-6=34 aligned

JEE Main 20257 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 7 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2025 · 7 April · Evening Shift · Mathematics · Binomial Theorem

The sum of the series 2 × 1 × ^ 20 C_4-3 × 2 × ^ 20 C_5+4 × 3 × ^...

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

Step-by-step 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}

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