====== Question 12 Exercise 7.3 ======
Solutions of Question 12 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.

Q12 If $2 y=\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2^4}+\frac{1.3 \cdot 5}{3 !} \cdot \frac{1}{2^6}+\ldots$ then show that $4 y^2+4 y-1=0$.
Solution: We are given
$$
2 y=\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2^4}-\frac{1.3 \cdot 5}{3 !} \cdot \frac{1}{2^6}+\ldots
$$

Adding 1 to both sides of the above equation, we get
$S=2 y+1=1+\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2^4}+$
$\frac{1.3 .5}{3 !} \cdot \frac{1}{2^6}+\ldots$
Now the above series is binomial series. Lel it be identical with the expansion of $(1+x)^n$ that is
$$
\begin{aligned}
& 10+n x+\frac{n(n-1)}{2 !} x^2+ \\
& \frac{n(n-1(n-2))}{3 !} x^3+\ldots
\end{aligned}
$$

Comparing both the series, we have $n x=\frac{1}{2^2}=\frac{1}{4}.... (1)$ 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
$$
\begin{aligned}
& \frac{n-1}{2 n}=\frac{1.3}{2 !} \cdot \frac{1}{2^4} \cdot 16=\frac{3}{2} \\
& \Rightarrow 6 n=2 n-2 \text { or } n=-\frac{1}{2} .
\end{aligned}
$$

Putting $n=-\frac{1}{2}$ in Eq.(1), we get $-\frac{1}{2} x=\frac{1}{4} \Rightarrow x=-\frac{1}{2}$. Thus
$$
\begin{aligned}
& 2 y+1=\left(1-\frac{1}{2}\right)^{-\frac{1}{2}} \\
& \Rightarrow 2 y+1=\left(\frac{1}{2}\right)^{-\frac{1}{2}} \\
& =\left(2^{-1}\right)^{-\frac{1}{2}}-\sqrt{2}
\end{aligned}
$$
$$
\frac{n(n-1)}{2 !} x^2=\frac{1.3}{2 !} \cdot \frac{1}{2^4}
$$

Taking square of Eq.(1), we have $n^2 x^2=\frac{1}{16}$
Dividing Eq.(2) by Eq.(3), we get
$$
\begin{aligned}
& \frac{n-1}{2 n}=\frac{1.3}{2 !} \cdot \frac{1}{2^4} \cdot 16=\frac{3}{2} \\
& \Rightarrow 6 n=2 n-2 \text { or } n=-\frac{1}{2} .
\end{aligned}
$$

Putting $n=-\frac{1}{2}$ in Eq.(1), we get $-\frac{1}{2} x=\frac{1}{4} \Rightarrow x=-\frac{1}{2}$. Thus
$$
\begin{aligned}
& 2 y+1=\left(1-\frac{1}{2}\right)^{-\frac{1}{2}} \\
& \Rightarrow 2 y+1=\left(\frac{1}{2}\right)^{-\frac{1}{2}} \\
& =\left(2^{-1}\right)^{-\frac{1}{2}}-\sqrt{2}
\end{aligned}
$$
Taking square of the both sides
$$
\begin{aligned}
& (2 y+1)^2=2 \\
& \Rightarrow 4 y^{+} 4 y+1=2 \\
& \Rightarrow 4 y^2+4 y-1=0 .
\end{aligned}
$$

Which is the required result.







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