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- Question 10 Exercise 7.2
- $. Solution: We know that $$ \left.(1+x)^n=\left(\begin{array}{l} n \\ \vdots \end{array}\right)+\left(\begin{array}{l} m \\ 1 \end{array}\right) x+\left(\begin{array}{l} n \\ 2 \end{array}\right) x^2-\ldots+i_n^*\... the above equation, we have $(1 \div 1)^n=\left(\begin{array}{l}n \\ 0\end{array}\right)+\left(\begin{ar
- Question 11 Exercise 7.1
- PTBB) Peshawar, Pakistan. =====Question 11===== \begin{align} & \left(\begin{array}{l} 2 \\ 2 \end{array}\right)+\left(\begin{array}{l} 3 \\ 2 \end{array}\right)+\left(\begin{array}{l} 4 \\ 2 \end{array}\right)+\ldots+\left(\begin
- Question 10 Exercise 7.1
- formulas below by mathematical induction, $\left(\begin{array}{1}5 \\5 \end{array}\right)+\left(\begin{array}{l}6 \\ 5\end{array}\right)+\left(\begin{array}{l}7 \\ 5\end{array}\right)+\ldots+\left(\begin{array}{c}n+4 \\ 5\end{array}\right)=\left(\begin{a
- Question 7 Exercise 7.2
- t{3})^5$ ====Solution==== Using binomial formula \begin{align}(2+\sqrt{3})^5+(2 \cdot \sqrt{3})^5& =[(2)^... \cdot(\sqrt{3})^5\end{align} simplifing, we get \begin{align} & =2 \cdot 2^5+2^5 C_2 \cdot 2^3 \cdot(\sq... 2})^{-}$ ====Solution==== Using binomial formula \begin{align} (1+\sqrt{2})^4-(1-\sqrt{2})^4 & =[1+{ }^4 ... \cdot(\sqrt{2})^4]\end{align} simplifing, we get \begin{align} & =2^4 C_1 \cdot \sqrt{2}+2^4 C_3 \cdot(\s
- Question 4 Exercise 7.2
- {r, 1}$ be the term containing $x^{23}$ that is: \begin{align}T_{r-1}&=\dfrac{20 !}{(20-r) ! r !}(x^2)^{2... _{r-1}$ containing $x^{33}$ is possibic only if \begin{align}x^{40 \cdot r}&=x^{23}\\ \Rightarrow 40-r&=... 7\end{align} Putting $r=17$. in $T_{r+1}$ we get \begin{align}T_{17-1}&=\dfrac{20 !}{(20 \cdot 17) ! 17 !... + 1}$ be the term containing $x^{23}$ that is: \begin{align}T_{r+1}&=\dfrac{8 !}{(8-r) ! r !}(2)^{8-r}(
- Question 10 Exercise 7.3
- tical with the expansion of $(1+x)^n$ that is $$ \begin{aligned} & 1+n x+\frac{n(n-1)}{2 !} x^2 \\ & +\fr... rac{1}{16}$ Dividing Eq.(2) by Eq.(3), we get $$ \begin{aligned} & \frac{n-1}{2 n}=\frac{3}{32} \cdot 16=... $ Putting $n=-\frac{1}{2}$ in Eq.(1), we get $$ \begin{aligned} & -\frac{1}{2} x=-\frac{1}{4} \\ & \Righ... tical with the expansion of $(1+x)^n$ that is $$ \begin{aligned} & 1+n x+\frac{n(n-1)}{2 !} x^2 \\ & +\fr
- Question 3 Exercise 7.2
- expansion. $T_{r+1}$ of the given expansion is: \begin{align}T_{r+1}&=\dfrac{9 !}{(9-r) ! r !}(\dfrac{4 ... rrow r=6 $$ Putting $r=6$ in the above $T_{r+1}$ \begin{align}T_{6-1}&=\dfrac{9 !}{(9-6) ! 6 !}\cdot \dfr... expansion. $T_{r+1}$ of the given expansion is: \begin{align} T_{r+1}&=\dfrac{10 !}{(10-r) ! r !}(x)^{10... tarrow r=2 $$ Putting $r=2$ in the above, we get \begin{align} T_{2+1}&=\dfrac{10 !}{(10-2) ! 2 !}(-3)^2
- Question 1 Exercise 7.3
- Using binomial theorem to tind the four terms $$ \begin{aligned} & (1-x)^{\frac{1}{2}}=1+\frac{1}{2} x+ \... c{1}{2}-1\right)}{2 !}(-x)^2 \end{aligned} $$ $$ \begin{aligned} & +\frac{-\frac{1}{2}\left(-\frac{1}{2}-... aligned} $$ Solution: Using binomial theorem $$ \begin{aligned} & (1-x)^{\frac{3}{2}}=1-\frac{3}{2} x+\f... $$ (iii) $(8+12 x)^{\frac{2}{3}}$ Solution: $$ \begin{aligned} & (8+12 x)^{\frac{2}{3}}=8^{\frac{2}{3}}
- Question 5 and 6 Exercise 7.3
- 3}-3 x j^{\frac{2}{3}}}{2 \cdot 3 x+4-5 x} $$ $$ \begin{aligned} & =\frac{8^{\frac{2}{3}}\left(1+\frac{3 ... and neglecting $x^2$ and higher powers of $x$ $$ \begin{aligned} & =\left(1-\frac{3 x}{2}+\frac{x}{4}+\te... cting $x^2$ and higher powers of $x$, we have $$ \begin{aligned} & =1+\frac{5 x}{8}-\frac{5 x}{x} \\ & =1... x^2-2 x}}{(x+1)^2} $$ Solution: We are given $$ \begin{aligned} & \frac{x \sqrt{x^2-2 x}}{(x+1)^2}=\frac
- Question 12 Exercise 7.3
- tical with the expansion of $(1+x)^n$ that is $$ \begin{aligned} & 10+n x+\frac{n(n-1)}{2 !} x^2+ \\ & \f... rac{1}{16}$ Dividing Eq.(2) by Eq.(3), we get $$ \begin{aligned} & \frac{n-1}{2 n}=\frac{1.3}{2 !} \cdot ... \frac{1}{4} \Rightarrow x=-\frac{1}{2}$. Thus $$ \begin{aligned} & 2 y+1=\left(1-\frac{1}{2}\right)^{-\fr... rac{1}{16}$ Dividing Eq.(2) by Eq.(3), we get $$ \begin{aligned} & \frac{n-1}{2 n}=\frac{1.3}{2 !} \cdot
- Question 2 Exercise 7.3
- laces. (i) $\sqrt{26}$ Solution: We are given $$ \begin{aligned} & \sqrt{26}=\sqrt{25+1} \\ & =\sqrt{25} ... }} \end{aligned} $$ Using binomial expansion $$ \begin{aligned} & \sqrt{26}=5\left[1+\frac{1}{25}\right]... } \text {. } $$ Using binomial expansion now $$ \begin{aligned} & =1-\frac{1}{2}(-0.002)+ \\ & \frac{-\f... olution: The cube root of 126 can be written as: \begin{aligned} & =5\left[1+\frac{1}{3} \cdot \frac{1}{1
- Question 12 Exercise 7.1
- 4} \in \mathbb{Z}$$ 3. For $n=k+1$ then consider \begin{align}\dfrac{5^{2(k+1)}-1}{24}&=\dfrac{5^{2 k+2}-... s an integer. ====Solution==== 1. For $n=1$ then \begin{align}\dfrac{10^{n+1}-9 n-10}{81}&=\dfrac{10^{i+1... ue for $n=1$. 2. Let it be true for $n=k$, then \begin{align}\dfrac{10^{k+1}-9 k-10}{81} \in \mathbb{Z}\... dots (i)\end{align} 3. For $n=k+1$, then we have \begin{align}\dfrac{10^{k+1+1}-9(k+1)-10}{81}& =\dfrac{1
- Question 2 Exercise 7.2
- a^r $$ For $4^{\text {th }}$ term, putting $r=3$ \begin{align} & T_{3+1}=\dfrac{7 !}{(7-3) ! 3 !} 2^{7-3}... })^r$$ For $8^{\text {th }}$ term, putting $r=7$ \begin{align}T_{7+1}&=\dfrac{10 !}{(10-7) ! 7 !}(\dfrac{... expansion. $T_{r+1}$ of the given expansion is: \begin{align}T_{r-1}&=\dfrac{21 !}{(21-r) ! r !}(x)^{21-... tarrow r=7 $$ Putting $r=7$ in the above, we get \begin{align}T_{7+1}&=\dfrac{21 !}{(21-7) ! 7 !}(-1)^7 \
- Question 5 Exercise 7.2
- (b x)^r$$ To get middle term $T_5$, we put $r=4$ \begin{align}T_5&=\dfrac{8 !}{(8-4) ! 4 !}(\dfrac{a}{x})... }{2})^r$$ Putting $r=4$ to get first middle term \begin{align} T_5&=\dfrac{9 !}{(9-4) ! 4 !}(3 x)^{9-4}(-... o get the $2^{\text {nd }}$ middle term that is: \begin{align}T_6&=\dfrac{9 !}{(9-5) ! 5 !}(3 x)^{9-5}(-\... rac{y}{3})^r$$ To get middle term, putting $r=5$ \begin{align}T_6&=\dfrac{10 !}{(10-5) ! 5 !}(3 x^2)^{10-
- Question 8 Exercise 7.2
- term in $\left(1-\frac{3 x}{2}\right)^{10}$ is: \begin{aligned} & \frac{(1+n) i}{1+i} \quad \frac{1+10}{... in $: 3-\mathbf{2}_1 1^{10}$ Now $T_{5} !=\left(\begin{array}{c}10 \\ 5\end{array}\right) 3^{10} 5-2 \gamma^{15}$ we compute at $x=\frac{3}{4}$, so $$ \begin{aligned} & T_{51}=\left(\begin{array}{c} 0 \\ 5 \end{array}\right) 3^{10 i-5} \quad 2, \frac{3}{4} y \\