JEE Main 2025 — 7 April, Evening Shift — Question 43

$(1+x)^{20}={ }^{20} C_{0}+{ }^{20} C_{1} x+{ }^{20} C_{2} x{ }^{2}+{ }^{20} C_{3} x{ }^{3} +{ }^{20} C_{4} x^{4}+{ }^{20} C_{5} x^{5}+\ldots .+{ }^{20} C_{20} x^{20}$

$$\begin{aligned} \frac{(1+x)^{20}}{x^{2}}=\frac{1}{x^{2}}+ & \frac{20}{x}+{ }^{20} C_{2}+{ }^{20} C_{3} x +{ }^{20} C_{4} x^{2}+{ }^{20} C_{5} x^{3}+\ldots+{ }^{20} C_{20} x^{18} \end{aligned}$$

d.w.r. to $x$

$$\begin{aligned} & \frac{x^{2} \cdot 20(1+x)^{19}+(1+x)^{20} \cdot 2 x}{x^{4}}=\frac{-2}{x^{3}}-\frac{20}{x^{2}}+{ }^{20} C_{3} +2 \cdot{ }^{20} C_{4} x+3 \cdot{ }^{20} C_{5} x^{2}+\ldots+18{ }^{20} C_{20} x^{17} \\& \frac{2 x(1+x)^{19}(11 x+1)}{x^{4}}=\frac{-2}{x^{3}}-\frac{20}{x^{2}}+{ }^{20} C_{3} +2 \cdot{ }^{20} C_{4} x+3 \cdot{ }^{20} C_{5} x^{2}+\ldots+18{ }^{20} C_{20} x^{17} \end{aligned}$$

Differentiate w.r. to $x$

$\frac{2\left[x^{3}(1+x)^{19} \cdot 11+x^{3}(11 x+1) \cdot 19(1+x)^{18}-(1+x)^{19}(11 x+1) \cdot 3 x^{2}\right]}{x^{6}}$

$=\frac{6}{x^{4}}+\frac{40}{x^{3}}+2 \cdot 1 \cdot{ }^{20} C_{4}+3 \cdot 2{ }^{20} C_{5} x+\ldots+18 \cdot 17{ }^{20} C_{20} x^{16}$

Put $x=-1$ in above equation

$\Rightarrow 0=6-40+R \Rightarrow R=34$