# Question 2 Exercise 7.3

Solutions of Question 2 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.

Q2 Find correct to three decimal places. (i) $\sqrt{26}$ Solution: We are given \begin{aligned} & \sqrt{26}=\sqrt{25+1} \\ & =\sqrt{25} \sqrt{1+\frac{1}{25}}=5\left[1+\frac{1}{25}\right]^{\frac{1}{2}} \end{aligned}

Using binomial expansion \begin{aligned} & \sqrt{26}=5\left[1+\frac{1}{25}\right]^{\frac{1}{2}} \\ & =5\left[1+\frac{1}{2} \cdot \frac{1}{25}+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2 !}\left(\frac{1}{25}\right)^2+\right. \\ & \left.\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3 !}\left(\frac{1}{25}\right)^3+\ldots\right] \\ & =5\left[1+\frac{1}{50}-\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{625}+\right. \\ & \left.\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} \cdot \frac{1}{15625}+\ldots\right] \\ & =5\left[1+\frac{1}{50}-\frac{1}{8 \times 625}+\right. \\ & \left.\frac{1}{16 \times 15625}+\ldots\right] \\ & =5[1+0.02-0.0002 \\ & +0.000008+\ldots] \\ & -5[1.019808 . .] \cong 5.099 \text {. } \\ & \end{aligned} (ii) $\frac{1}{\sqrt{0.998}}$

Solution: We are given that $$\frac{1}{\sqrt{0.998}}=(0.998)^{-\frac{1}{2}}$$ $$=(1-0.002)^{-\frac{1}{2}} \text {. }$$

Using binomial expansion now \begin{aligned} & =1-\frac{1}{2}(-0.002)+ \\ & \frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)}{2 !}(-0.002)^2+ \\ & \frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3 !}(-0.002)^3 \\ & +\ldots \ldots \\ & =1+0.001+\frac{1}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot(0.000004)+ \\ & \frac{1}{2} \cdot \frac{3}{2} \cdot \frac{5}{2}(0.000000008)+\ldots \\ & =1+0.001+0.0000015+ \\ & 0.000000005+\ldots \\ & \cong 1.001 . \end{aligned} (iii) Find cube root of 126 correct to five decimal places. Solution: The cube root of 126 can be written as: \begin{aligned} & =5\left[1+\frac{1}{3} \cdot \frac{1}{125}+\right.
& \frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2 !}\left(\frac{1}{125}\right)^2+
& \left.\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)\left(\frac{1}{3}-2\right)}{3 !}\left(\frac{1}{125}\right)^3+\ldots\right]
& =5\left[1+\frac{1}{3 \times 125}-\frac{1}{3} \cdot \frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{15625}+\right. \end{aligned} \begin{aligned} & \left.\frac{1}{3} \cdot \frac{2}{3} \cdot \frac{5}{3} \cdot \frac{1}{6} \cdot \frac{1}{1953125}+\ldots\right]
& =5\left[1+\frac{1}{375}-\frac{1}{9 \times 15625}+\right.
& \left.\frac{5}{27 \times 1953125}+\ldots\right]
& =5[1+0.0026667-0.00000711+
& 0.00000009482+\ldots]
& \approx 5(1.00265968482)
& \approx 5.013298 . \end{aligned}