JEE Main 2025 Apr 7 Shift 2, Mathematics Q25: Properties Of Binomial Coefficient

Let

$(1-x)^{20} = {}^{20}C_0 -{}^{20}C_1x +{}^{20}C_2x^2 -\cdots +{}^{20}C_{20}x^{20}$

Dividing by $x^2$,

$\dfrac{(1-x)^{20}}{x^2} = \dfrac{{}^{20}C_0}{x^2} -\dfrac{{}^{20}C_1}{x} +{}^{20}C_2 -{}^{20}C_3x +{}^{20}C_4x^2 -\cdots$

Differentiate both sides twice and substitute $x=1$.

Since the left-hand side becomes $0$,

$6-2{}^{20}{C}_1+S=0$

$\Rightarrow S=2\times 20-6=34$

$\boxed{34}$