Question 1 Exercise 7.3
Solutions of Question 1 of Exercise 7.3 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
QI Find the first four terms in the expansion of (i) (1 - x) $\frac{1}{2}$ Solution: Using binomial theorem to tind the four terms $$ \begin{aligned} & (1-x)^{\frac{1}{2}}=1+\frac{1}{2} x+ \\ & \frac{\frac{1}{2}\left(-\frac{1}{2}-1\right)}{2 !}(-x)^2 \end{aligned} $$ $$ \begin{aligned} & +\frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3 !}(-x)^3+. . \\ & =1+\frac{x}{2}+\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdot \frac{1}{2} x^2+ \\ & \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right) \cdot \frac{1}{24} \cdot(-x)^3+\ldots \ldots \\ & =1+\frac{x}{2}+\frac{3 x^2}{8}+\frac{5 x^3}{16}+\ldots \\ & \text { (ii) }(1-x)^{\frac{3}{2}} \\ & (1-x)^{\frac{3}{2}}=1-\frac{3}{2} x+\frac{\frac{3}{2}\left(\frac{3}{2}-1\right)}{2 !}(-x)^2 \\ & +\frac{\frac{3}{2}\left(\frac{3}{2}-1\right)\left(\frac{3}{2}-2\right)}{3 !}(-x)^3+\ldots \\ & =1-\frac{3}{2} x+\frac{3}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} x^2+ \\ & \frac{3}{2} \cdot \frac{1}{2} \cdot\left(-\frac{1}{2}\right) \cdot \frac{1}{6}(-x)^3+\ldots \\ & (1-x)^{\frac{3}{2}}=1-\frac{3}{2} x+\frac{3}{8} x^2+\frac{1}{16} x^3+\ldots \\ & \end{aligned} $$
Solution: Using binomial theorem $$ \begin{aligned} & (1-x)^{\frac{3}{2}}=1-\frac{3}{2} x+\frac{\frac{3}{2}\left(\frac{3}{2}-1\right)}{2 !}(-x)^2 \\ & +\frac{\frac{3}{2}\left(\frac{3}{2}-1\right)\left(\frac{3}{2}-2\right)}{3 !}(-x)^3+\ldots \end{aligned} $$ (iii) $(8+12 x)^{\frac{2}{3}}$
Solution: $$ \begin{aligned} & (8+12 x)^{\frac{2}{3}}=8^{\frac{2}{3}}\left(1+\frac{12 x}{8}\right)^{\frac{2}{3}} \\ & =\left(2^3\right)^{\frac{2}{3}}\left(1+\frac{3 x}{2}\right)^{\frac{2}{3}}=4\left(1+\frac{3 x}{2}\right)^{\frac{2}{3}} \end{aligned} $$
Now using binomal expansion $$ \begin{aligned} & =4\left[1+\frac{2}{3} \cdot \frac{3 x}{2}+\frac{\frac{2}{3}\left(\frac{2}{3}-1\right)}{2 !}\left(\frac{3 x}{2}\right)^2\right. \\ & \left.+\frac{\frac{2}{3}\left(\frac{2}{3}-1\right)\left(\frac{2}{3}-2\right)}{3 !}\left(\frac{3 x}{2}\right)^3+\ldots\right] \\ & =4\left[1+x-\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{19 x^2}{2}+\right. \end{aligned} $$ $$ \begin{aligned} & \left.\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{4}{3} \cdot \frac{27 x^3}{8}+\ldots\right] \\ & =4\left[1+x-\frac{x^2}{4}++\frac{x^3}{6}-\ldots\right] \end{aligned} $$
Go To