Question 10 Exercise 7.3

Solutions of Question 10 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.

Q10 Find the sum of the following series: (i) $1-\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2^4}+\ldots$ Solution: The given series is binomial series. Let it be identical with the expansion of $(1+x)^n$ that is $$\begin{aligned} & 1+n x+\frac{n(n-1)}{2 !} x^2 \\ & +\frac{n(n-1(n-2))}{3 !} x^3+\ldots \end{aligned}$$

Comparing both the series, we have $n x=-\frac{1}{4}$ (I) and $\frac{n(n-1)}{2 !} x^2=\frac{1.3}{2 !} \cdot \frac{1}{2^4}$ Taking square of Eq.(1), we have $n^2 x^2=\frac{1}{16}$ Dividing Eq.(2) by Eq.(3), we get $$\begin{aligned} & \frac{n-1}{2 n}=\frac{3}{32} \cdot 16=\frac{3}{2} \\ & \Rightarrow 3 n=n-1 \Rightarrow n=-\frac{1}{2} . \end{aligned}$$

Putting $n=-\frac{1}{2}$ in Eq.(1), we get $$\begin{aligned} & -\frac{1}{2} x=-\frac{1}{4} \\ & \Rightarrow x=\frac{1}{2} . \text { Thus } \\ & \left(1+\frac{1}{2}\right)^{-\frac{1}{2}}=1-\frac{1}{2^2}+\frac{1 \cdot 3}{2 !} \cdot \frac{1}{2^4}+\ldots \\ & \Rightarrow\left(\frac{3}{2}\right)^{\frac{1}{2}}=1-\frac{1}{2^2}+\frac{1 \cdot 3}{2 !} \cdot \frac{1}{2^4}+\ldots \\ & \Rightarrow \sqrt{\frac{2}{3}}=1-\frac{1}{2^2}+\frac{1 \cdot 3}{2 !} \cdot \frac{1}{2^4}+\ldots \end{aligned}$$

Hence the sum of the series is $\sqrt{\frac{2}{3}}$. (ii) $1+\frac{5}{8}+\frac{5.8}{8.12}+\frac{5.8 .11}{8.12 .16}+\ldots$

Solution: The given series is binomial series. Let it be identical with the expansion of $(1+x)^n$ that is $$\begin{aligned} & 1+n x+\frac{n(n-1)}{2 !} x^2 \\ & +\frac{n(n-1(n-2))}{3 !} x^3+\ldots \end{aligned}$$ Comparing both the series, we have $$\begin{aligned} & n x=\frac{5}{8} \\ & \frac{n(n-1)}{2 !} x^2=\frac{5.8}{8.12} . \end{aligned}$$

Taking square of Eq.(1), we have $$n^2 x^2=\frac{25}{64}$$

Dividing Eq.(2) by Eq.(3), we get $$\begin{aligned} & \frac{n-1}{2 n}=\frac{5.8}{8.12} \cdot \frac{64}{25}=\frac{16}{15} \\ & \Rightarrow 32 n=15 n-15 \\ & \Rightarrow 32 n-15 n=-15 \\ & \Rightarrow 17 n=-15 \text { or } n=-\frac{15}{17} . \end{aligned}$$

Putting $n=\frac{15}{17}$ in Eq.(1), we get $$-\frac{15}{17} x=\frac{5}{8}$$ $\Rightarrow x=-\frac{17}{24}$. Hence $$\begin{aligned} & \left(1-\frac{17}{24}\right)^{\frac{15}{17}}=1+\frac{5}{8}+\frac{5.8}{8.12} \\ & +\frac{5.8 .11}{8.12 .16}+\ldots \\ & \Rightarrow\left(\frac{24}{7}\right)^{\frac{15}{17}}=1+\frac{5}{8}+\frac{5.8}{8.12} \\ & +\frac{5.8 .11}{8.12 .16}+\ldots . \end{aligned}$$

Hence the sum of the given series is: $$\left(\frac{24}{7}\right)^{\frac{15}{17}}$$

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