Question

The value of ${}^{30}{C}_0-\frac{{}^{30}{C}_1}{2}+\frac{{}^{30}{C}_2}{3}-\cdots \cdots +\frac{{}^{30}{C}_{30}}{31}$ is

• A. $\frac{1}{30}$
• B. $\frac{30}{31}$
• C. $\frac{31}{30}$
• D. $\frac{1}{31}$

Answer 1

The correct option is D $\frac{1}{31}$

$(1-x{)}^{30}={}^{30}{C}_0-{}^{30}{C}_1\cdot x+{}^{30}{C}_2\cdot x^2-\cdots +{}^{30}{C}_{30}\cdot x^{30}$
Integrating both sides w.r.t. $x$ from $x=0$ to $x=1$
$\underset{0}{\overset{1}{\int }}(1-x{)}^{30}dx=\underset{0}{\overset{1}{\int }}{}^{30}{C}_{0}dx-\underset{0}{\overset{1}{\int }}{}^{30}{C}_{1}xdx+\underset{0}{\overset{1}{\int }}{}^{30}{C}_2{x}^{2}dx-\cdots +\underset{0}{\overset{1}{\int }}{}^{30}{C}_{30}{x}^{30}dx$
$\Rightarrow {[-\frac{(1-x{)}^{31}}{31}]}_0^1={\left[{}^{30}{C}_{0}x\right]}_0^1-{\left[{}^{30}{C}_1\frac{x^2}{2}\right]}_0^1+{\left[{}^{30}{C}_2\frac{x^3}{3}\right]}_0^1-\cdots +{\left[{}^{30}{C}_{30}\frac{x^{31}}{31}\right]}_0^1$
Putting $x=1$, we get
${}^{30}{C}_0-\frac{{}^{30}{C}_1}{2}+\frac{{}^{30}{C}_2}{3}-......+\frac{{}^{30}{C}_{30}}{31}=\frac{1}{31}$