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- Question 10 Exercise 7.2
- ^n=\left(\begin{array}{l} n \\ \vdots \end{array}\right)+\left(\begin{array}{l} m \\ 1 \end{array}\right) x+\left(\begin{array}{l} n \\ 2 \end{array}\right) x^2-\ldots+i_n^*\right) x^n \cdot $$ Putting $x=1$ in the above equation, we have $(1 \div 1)^n=\left(
- Question 5 and 6 Exercise 7.3
- ed} & =\frac{8^{\frac{2}{3}}\left(1+\frac{3 x}{8}\right)^{\frac{2}{3}}}{2\left(1+\frac{3 x}{2}\right) \sqrt{4}\left(1-\frac{5 x}{4}\right)^{\frac{1}{2}}} \\ & =\frac{\left(2^3\right)^{\frac{2}{3}}}{2.2}\left[\left(1+\frac{3 x}{8}\right)^{\fra
- Question 10 Exercise 7.1
- duction, $\left(\begin{array}{1}5 \\5 \end{array}\right)+\left(\begin{array}{l}6 \\ 5\end{array}\right)+\left(\begin{array}{l}7 \\ 5\end{array}\right)+\ldots+\left(\begin{array}{c}n+4 \\ 5\end{array}\right)=\left(\begin{array}{c}n+5 \\ 6\end{array}\right)
- Question 1 Exercise 7.3
- {2} x+ \\ & \frac{\frac{1}{2}\left(-\frac{1}{2}-1\right)}{2 !}(-x)^2 \end{aligned} $$ $$ \begin{aligned} & +\frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3 !}(-x)^3+. . \\ & =1+\frac{x}{2}+\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdot \frac{1}{2} x^2+
- Question 13 Exercise 7.3
- frac{1}{n} x+\frac{\frac{1}{n}\left(\frac{1}{n}-1\right)}{2 !} x^2+\ldots \end{aligned} $$ We are conside... n+(n-1) x}=\frac{2 n\left[1+\left(\frac{n+1}{2 n}\right) x\right]}{2 n\left[1+\left(\frac{n-1}{2 n}\right) x\right]} \\ & =\left[1+\left(\frac{n+1}{2 n}\right) x\right]
- Question 2 Exercise 7.3
- t{25} \sqrt{1+\frac{1}{25}}=5\left[1+\frac{1}{25}\right]^{\frac{1}{2}} \end{aligned} $$ Using binomial e... \begin{aligned} & \sqrt{26}=5\left[1+\frac{1}{25}\right]^{\frac{1}{2}} \\ & =5\left[1+\frac{1}{2} \cdot \frac{1}{25}+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2 !}\left(\frac{1}{25}\right)^2+\right. \\ & \left.\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\f
- Question 7 Exercise 7.2
- &=\left[\left(\begin{array}{l} 5 \\ 0 \end{array}\right) a^5+\left(\begin{array}{l} 5 \\ 1 \end{array}\right) a^4 b\right. \\ & +\left(\begin{array}{l} 5 \\ 2 \end{array}\right) a^3 b^2+\left(\begin{array}{l} 5 \\ 3 \end{array
- Question 11 Exercise 7.1
- lign} & \left(\begin{array}{l} 2 \\ 2 \end{array}\right)+\left(\begin{array}{l} 3 \\ 2 \end{array}\right)+\left(\begin{array}{l} 4 \\ 2 \end{array}\right)+\ldots+\left(\begin{array}{l} n \\ 2 \end{array}\right)=\left(\begin{array}{c} n+1 \\ 3 \end{array}\righ
- Question 10 Exercise 7.3
- n-1}{2 n}=\frac{3}{32} \cdot 16=\frac{3}{2} \\ & \Rightarrow 3 n=n-1 \Rightarrow n=-\frac{1}{2} . \end{aligned} $$ Putting $n=-\frac{1}{2}$ in Eq.(1), we get $... egin{aligned} & -\frac{1}{2} x=-\frac{1}{4} \\ & \Rightarrow x=\frac{1}{2} . \text { Thus } \\ & \left(1+\frac{1}{2}\right)^{-\frac{1}{2}}=1-\frac{1}{2^2}+\frac{1 \cdot 3}{
- Question 3 Exercise 7.3
- -\frac{x}{2}+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2 !}(-x)^2+\right. \\ & \left.\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3 !}(-x)^3+\ldots\right] \times \\ & {\left[1-\frac{x}{2}+\frac{-\frac{1}{2}
- Question 7 and 8 Exercise 7.3
- +\frac{x}{4}+\frac{\frac{1}{4}\left(\frac{1}{4}-1\right)}{2 !} x^2+\right. \\ & \left.\frac{\frac{1}{4}\left(\frac{1}{4}-1\right)\left(\frac{1}{4}-2\right)}{3 !} x^3+\ldots\right] \times \\ & 1-\frac{x}{4}+\frac{\frac{1}{4}\left(\fra
- Question 12 Exercise 7.3
- !} \cdot \frac{1}{2^4} \cdot 16=\frac{3}{2} \\ & \Rightarrow 6 n=2 n-2 \text { or } n=-\frac{1}{2} . \end... }$ in Eq.(1), we get $-\frac{1}{2} x=\frac{1}{4} \Rightarrow x=-\frac{1}{2}$. Thus $$ \begin{aligned} & 2 y+1=\left(1-\frac{1}{2}\right)^{-\frac{1}{2}} \\ & \Rightarrow 2 y+1=\left(\frac{1}{2}\right)^{-\frac{1}{2}} \\ & =\left(2^{-1}\right)
- Question 3 Exercise 7.2
- nt of $x$ is possible only if $x^{18-3 r}=x^0$ $$\Rightarrow 18-3 r=0 \Rightarrow 3 r=18 \Rightarrow r=6 $$ Putting $r=6$ in the above $T_{r+1}$ \begin{align}T_{6-1}&=\dfrac{9 !}{(9... ^3} \cdot \dfrac{3^6}{2^6} \because(-3)^6=3^6 \\ \Rightarrow T_7&=84 \cdot \dfrac{2^6}{2^6} \cdot 3^{6-3}
- Question 9 Exercise 7.3
- of $x^{\prime \prime}$ in $\left(\frac{1+x}{1-x}\right)^2$. Solution: Given that: $$ \begin{aligned} & \left(\frac{1+x}{1-x}\right)^2=(1+x)^2(1-x)^{-2} \\ & =\left(x^2+2 x+1\right)(1-x)^2 \end{aligned} $$ Applying binomial theorem $$ \begin{aligned} & =\left(x^2+2 x+1\right)[1+2 x+ \\ & \frac{-2(-2-1)}{2 !}(-x)^2 \\ & \lef
- Question 9 Exercise 7.2
- $ \begin{aligned} & \left(x \quad y=20(12-y)^{20}\right. \\ & =12^{2 n}\left(\begin{array}{ll} 1 & \frac{y}{12} \end{array}\right)^{31} \end{aligned} $$ Now we theck $\frac{(n+1) \cdot x}{1+|x|}$ for $\left(\frac{1}{12}\right)^2 \cdot$ Here $n=20$ and $\left.\left|x^{\prime}=\right|-\frac{y}{12} \right\rvert\,=\frac{1}{3}$ for $y=