Question 13 Exercise 7.3

Solutions of Question 13 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.

Q13 If $x$ is so small that $x^3$ and higher of $x$ can be ignored. Show that $n^{\text {th }}$ root of $1+x$ is equal to $\frac{2 n+(n+1) x}{2 n+(n-1) x}$ Solution: We have show that $$\begin{aligned} & (1+x)^{\frac{1}{n}}=\frac{2 n+(n+1) x}{2 n+(n-1) x} \\ & \frac{2 n+(n+1) x}{2 n+(n-1) x} \\ & =1+\frac{1}{n} x+\frac{\frac{1}{n}\left(\frac{1}{n}-1\right)}{2 !} x^2+\ldots \end{aligned}$$ We are considering $$\begin{aligned} & \frac{2 n+(n+1) x}{2 n+(n-1) x}=\frac{2 n\left[1+\left(\frac{n+1}{2 n}\right) x\right]}{2 n\left[1+\left(\frac{n-1}{2 n}\right) x\right]} \\ & =\left[1+\left(\frac{n+1}{2 n}\right) x\right] \times\left[1+\left(\frac{n-1}{2 n}\right) x\right]^{-1} \end{aligned}$$

Applying binomial theorem now $$\begin{aligned} & {\left[1+\left(\frac{n+1}{2 n}\right) x\right] \times} \\ & {\left[1-\frac{n-1}{2 n} x+\frac{-1(-1-1)}{2 !}\left(\frac{n-1}{2 n} x\right)^2\right.} \\ & +\ldots] \\ & =\left[1+\left(\frac{n+1}{2 n}\right) x\right] \times\left[1-\frac{n-1}{2 n} x+\right. \\ & \left.\left(\frac{n-1}{2 n}\right)^2 x^2+\ldots\right] \end{aligned}$$

Multiplying and taking tems upto $x^2$ $$\begin{aligned} & =1-\frac{n-1}{2 n} x+\left(\frac{n-1}{2 n}\right)^2 x^2+\ldots \\ & +\left(\frac{n+1}{2 n}\right) x-\left(\frac{n+1}{2 n}\right) \cdot \frac{n-1}{2 n} x^2+\ldots \\ & =1-\left(\frac{n-1}{2 n}-\frac{n+1}{2 n}\right) x+ \\ & \left(\frac{(n-1)^2}{4 n^2}-\frac{n^2-1}{4 n^2}\right) x^2+\ldots \\ & =1-\left(\frac{n-1-n-1}{2 n}\right) x+ \\ & \left(\frac{n^2-2 n+1-n^2+1}{4 n^2}\right) x^2+\ldots \\ & =1+\frac{2 n}{n} x+\frac{2-2 n}{2 n^2} \cdot \frac{1}{2} x^2+\ldots \\ & =1+\frac{1}{n} x+\frac{1-n}{n^2} \cdot \frac{1}{2} x^2+\ldots \\ & =1+\frac{1}{n} x+\frac{1}{n}\left(\frac{1}{n}-1\right) \cdot \frac{1}{2} x^2+\ldots \\ & \Rightarrow(1+x)^{\frac{1}{n}}=1+\frac{1}{n} x+\frac{\frac{1}{n}\left(\frac{1}{n}-1\right)}{2 !} x^2 \end{aligned}$$ $$+\ldots=\frac{2 n+(n+1) x}{2 n+(n-1) x} \text {. }$$

Which is the required result.

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