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- Question 10 Exercise 7.2
- osficient $s=2^{n-1}$. Solution: We know that $$ \left.(1+x)^n=\left(\begin{array}{l} n \\ \vdots \end{array}\right)+\left(\begin{array}{l} m \\ 1 \end{array}\right) x+\left(\begin{array}{l} n \\ 2 \end{array}\right) x^2-\ldot
- Question 5 and 6 Exercise 7.3
- x} $$ $$ \begin{aligned} & =\frac{8^{\frac{2}{3}}\left(1+\frac{3 x}{8}\right)^{\frac{2}{3}}}{2\left(1+\frac{3 x}{2}\right) \sqrt{4}\left(1-\frac{5 x}{4}\right)^{\frac{1}{2}}} \\ & =\frac{\left(2^3\right)^{\frac{2}{3}}}{2.2}\left[\left(1+\frac
- Question 10 Exercise 7.1
- h the formulas below by mathematical induction, $\left(\begin{array}{1}5 \\5 \end{array}\right)+\left(\begin{array}{l}6 \\ 5\end{array}\right)+\left(\begin{array}{l}7 \\ 5\end{array}\right)+\ldots+\left(\begin{array}{c}n+4 \\ 5\end{array}\right)=\left(
- Question 1 Exercise 7.3
- ac{1}{2}}=1+\frac{1}{2} x+ \\ & \frac{\frac{1}{2}\left(-\frac{1}{2}-1\right)}{2 !}(-x)^2 \end{aligned} $$ $$ \begin{aligned} & +\frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3 !}(-x)^3+. . \\ & =1+\frac{x}{2}+\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdo
- Question 2 Exercise 7.3
- qrt{25+1} \\ & =\sqrt{25} \sqrt{1+\frac{1}{25}}=5\left[1+\frac{1}{25}\right]^{\frac{1}{2}} \end{aligned}... nomial expansion $$ \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}\righ
- Question 13 Exercise 7.3
- +(n-1) x} \\ & =1+\frac{1}{n} x+\frac{\frac{1}{n}\left(\frac{1}{n}-1\right)}{2 !} x^2+\ldots \end{aligne... gned} & \frac{2 n+(n+1) x}{2 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(
- Question 7 Exercise 7.2
- mial theorem $$ \begin{aligned} (a+b)^5+(a-b)^5&=\left[\left(\begin{array}{l} 5 \\ 0 \end{array}\right) a^5+\left(\begin{array}{l} 5 \\ 1 \end{array}\right) a^4 b\right. \\ & +\left(\begin{array}{l} 5 \\ 2 \end{array}\right) a^3 b^
- Question 11 Exercise 7.1
- Pakistan. =====Question 11===== \begin{align} & \left(\begin{array}{l} 2 \\ 2 \end{array}\right)+\left(\begin{array}{l} 3 \\ 2 \end{array}\right)+\left(\begin{array}{l} 4 \\ 2 \end{array}\right)+\ldots+\left(\begin{array}{l} n \\ 2 \end{array}\right)=\left(
- Question 3 Exercise 7.3
- } & (1-x)^{\frac{1}{2}}(1+x)^{\frac{1}{2}} \\ & =\left[1-\frac{x}{2}+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2 !}(-x)^2+\right. \\ & \left.\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3 !}(-x)^3+\ldots\right] \times
- Question 7 and 8 Exercise 7.3
- & (1+x)^{\frac{1}{4}}+(1-x)^{\frac{1}{4}} \\ & =\left[1+\frac{x}{4}+\frac{\frac{1}{4}\left(\frac{1}{4}-1\right)}{2 !} x^2+\right. \\ & \left.\frac{\frac{1}{4}\left(\frac{1}{4}-1\right)\left(\frac{1}{4}-2\right)}{3 !} x^3+\ldots\right] \times \\
- Question 9 Exercise 7.3
- Find the coefficient of $x^{\prime \prime}$ in $\left(\frac{1+x}{1-x}\right)^2$. Solution: Given that: $$ \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 !}
- Question 9 Exercise 7.2
- 12$, then the give becounes $$ \begin{aligned} & \left(x \quad y=20(12-y)^{20}\right. \\ & =12^{2 n}\left(\begin{array}{ll} 1 & \frac{y}{12} \end{array}\right... Now we theck $\frac{(n+1) \cdot x}{1+|x|}$ for $\left(\frac{1}{12}\right)^2 \cdot$ Here $n=20$ and $\left.\left|x^{\prime}=\right|-\frac{y}{12} \right\rvert\
- Question 8 Exercise 7.2
- =\frac{3}{4}$. Solution: We first write in form $\left(3-2,1^{10}=3^{10}\left(1-\frac{3 x}{2}\right)^{10}\right.$. The numerically greatest term in $\left(1-\frac{3 x}{2}\right)^{10}$ is: \begin{aligned} ... reates in $: 3-\mathbf{2}_1 1^{10}$ Now $T_{5} !=\left(\begin{array}{c}10 \\ 5\end{array}\right) 3^{10}
- Question 12 Exercise 7.3
- x=-\frac{1}{2}$. Thus $$ \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} $$ $$... x=-\frac{1}{2}$. Thus $$ \begin{aligned} & 2 y+1=\left(1-\frac{1}{2}\right)^{-\frac{1}{2}} \\ & \Rightar
- Question 10 Exercise 7.3
- \Rightarrow x=\frac{1}{2} . \text { Thus } \\ & \left(1+\frac{1}{2}\right)^{-\frac{1}{2}}=1-\frac{1}{2^... {2 !} \cdot \frac{1}{2^4}+\ldots \\ & \Rightarrow\left(\frac{3}{2}\right)^{\frac{1}{2}}=1-\frac{1}{2^2}+... ow x=-\frac{17}{24}$. Hence $$ \begin{aligned} & \left(1-\frac{17}{24}\right)^{\frac{15}{17}}=1+\frac{5}... +\frac{5.8 .11}{8.12 .16}+\ldots \\ & \Rightarrow\left(\frac{24}{7}\right)^{\frac{15}{17}}=1+\frac{5}{8}