Haryana Police Constable Model Paper 4

CONCEPT:Binomial Theorem: According to the theorem, it is possible to expand any non-negative power of x+ y into a sum of the form: $(x+y)^{n}={}^{n}C_{0}x^{n}y^{0}+{}^{n}C_{1}x^{n-1}y^{1}+{}^{n}C_{2}x^{n-2}y^{2}+{}^{n}C_{3}x^{n-3}y^{3}+\dots+{}^{n}C_{n}x^{0}y^{n}$ where $n \geq 0$ is an integer. CALCULATION: $(1+x)^{50}=(1+{}^{50}C_{1}x^{1}+{}^{50}C_{2}x^{2}+{}^{50}C_{3}x^{3}+{}^{50}C_{4}x^{4}+\dots+{}^{50}C_{50}x^{50})$ ......equation (1) The above equations is written with the help of “BINOMIAL THEOREM” We need to find the sum of $\left({}^{50}C_{1}+{}^{50}C_{3}+{}^{50}C_{5}+\dots+{}^{50}C_{49}\right)$ , as they are the coefficients of odd powers of $x$ . In the equation (1), substitute x = 1 $\Rightarrow(1+1)^{50}=1+{}^{50}C_{1}+{}^{50}C_{2}+{}^{50}C_{3}+{}^{50}C_{4}+\dots+{}^{50}C_{50}$ $\Rightarrow2^{50}=1+{}^{50}C_{1}+{}^{50}C_{2}+{}^{50}C_{3}+{}^{50}C_{4}+\dots+{}^{50}C_{50} \dots$ equation (2) In the equation(1), substitute x = (-1) $\Rightarrow(1-1)^{50}=1-{}^{50}C_{1}+{}^{50}C_{2}-{}^{50}C_{3}+{}^{50}C_{4}+\dots+{}^{50}C_{50}$ $\Rightarrow0=1-{}^{50}C_{1}+{}^{50}C_{2}-{}^{50}C_{3}+{}^{50}C_{4}+\dots+{}^{50}C_{50} \dots$equation (3) By subtracting equation (3) from equation (2), $\Rightarrow2^{50}=2 \times \left({}^{50}C_{1}+{}^{50}C_{3}+{}^{50}C_{5}+{}^{50}C_{7}+\dots+{}^{50}C_{49}\right)$ $\Rightarrow{}^{50}C_{1}+{}^{50}C_{3}+{}^{50}C_{5}+{}^{50}C_{7}+\dots+{}^{50}C_{49}=2^{49}$ The sum of the coefficients of odd powers of $x = 2^{49}$