====== 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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