Mathematics - Binomial Theorem Question with Solution | TestHub

$\begin{array}{rl}& (1-\mathrm{x}{)}^{20}={}^{20}{\mathrm{C}}_0-{}^{20}{\mathrm{C}}_1\mathrm{x}+{}^{20}{\mathrm{C}}_2{\mathrm{x}}^2\dots ..+{}^{20}{\mathrm{C}}_{20}{\mathrm{x}}^{20}\\ & \frac{(1-\mathrm{x}{)}^{20}}{{\mathrm{x}}^2}=\frac{{}^{20}{\mathrm{C}}_0}{{\mathrm{x}}^2}-\frac{{}^{20}{\mathrm{C}}_1}{\mathrm{x}}+{}^{20}{\mathrm{C}}_2-{}^{20}{\mathrm{C}}_3\mathrm{x}+{}^{20}{\mathrm{C}}_4{\mathrm{x}}^2\dots \end{array}$ Diff twice and put $\mathrm{x}=1$ $\begin{array}{rl}& =6-{}^{20}{\mathrm{C}}_1(2)+\mathrm{A}\\ & \mathrm{A}=40-6=34\end{array}$