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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