====== Question 11 Exercise 7.3 ======
Solutions of Question 11 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.

Q11 If $y=\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2^4}+\frac{1 \cdot 3 \cdot 5}{3 !} \cdot \frac{1}{2^6}+\ldots$ then show that $y^2+2 y-1=0$.
Solution: We are given $y=\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2^4}+\frac{1.3 \cdot 5}{3 !} \cdot \frac{1}{2^6}+\ldots$
Adding 1 to both sides of the above equation
$$
S=y+1=1+\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2^4}
$$
$$
+\frac{1.3 .5}{3 !} \cdot \frac{1}{2^6}+\ldots
$$

Now the above series is binomial series. Let it be identical with the exparsion of $(1+x)^{\prime \prime}$ that is $1+n x=\frac{n(n-1)}{2 !} x^2+$ $\frac{n(n-1(n-2))}{3 !} x^3+\ldots$
Comparing both the series, we have $n x=\frac{1}{2^2}=\frac{1}{4}$ (1) 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
$$
\begin{aligned}
& \frac{n-1}{2 n}=\frac{1 \cdot 3}{2 !} \cdot \frac{1}{2^4} \cdot 16=\frac{3}{2} \\
& \Rightarrow 6 n=2 n-2 \text { or } n=-\frac{1}{2} . \\
& \text { Putting } n=-\frac{1}{2} \text { in Eq.(1), we get } \\
& -\frac{1}{2} x=\frac{1}{4} \Rightarrow x=-\frac{1}{2} \text {. Thus } \\
& y+1=\left(1-\frac{1}{2}\right)^{\frac{1}{2}}=\left(\frac{1}{2}\right)^{-\frac{1}{2}} \text { or } \\
& y+1=\left(2^{-1}\right)^{\frac{1}{2}}=\sqrt{2}
\end{aligned}
$$

Taking square of the both sides
$$
\begin{aligned}
& (y+1)^2=2 \\
& \Rightarrow y^2+2 y+1-2=0 \\
& \Rightarrow y^2+2 y-1=0 .
\end{aligned}
$$

Which is the desired result.




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