Question

The value of $\frac{{}^{50}{C}_0}{3}-\frac{{}^{50}{C}_1}{4}+\frac{{}^{50}{C}_2}{5}+\cdots +\frac{{}^{50}{C}_{50}}{53}$ is equal to

• A. ${\int }_0^1{x}^3(1-x{)}^{50}dx$
• B. ${\int }_0^{1}x(1-x{)}^{50}dx$
• C. $\frac{1}{51}-\frac{2}{52}+\frac{1}{53}$
• D. $\frac{1}{70278}$

Answer 1

Answer: (C), (D)

The correct options are
C

$\frac{1}{51}-\frac{2}{52}+\frac{1}{53}$

D

$\frac{1}{70278}$

$\frac{{}^{50}{C}_0}{3}-\frac{{}^{50}{C}_1}{4}+\cdots +\frac{{}^{50}{C}_{50}}{53}={\int }_0^1{}^{50}{C}_0{x}^2-{}^{50}{C}_1{x}^3+{}^{50}{C}_2{x}^4+\cdots +{}^{50}{C}_{50}{x}^{52}dx={\int }_0^1{x}^2(1-x{)}^{50}dx={\int }_0^1(1-x{)}^2{x}^{50}dx={\int }_0^1(1-2x+x^2)x^{50}dx=\frac{1}{51}-\frac{2}{52}+\frac{1}{53}$