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- Question 10 Exercise 7.2
- $. Solution: We know that $$ \left.(1+x)^n=\left(\begin{array}{l} n \\ \vdots \end{array}\right)+\left(\begin{array}{l} m \\ 1 \end{array}\right) x+\left(\begin{array}{l} n \\ 2 \end{array}\right) x^2-\ldots+... the above equation, we have $(1 \div 1)^n=\left(\begin{array}{l}n \\ 0\end{array}\right)+\left(\begin
- Question 10 Exercise 7.1
- an. =====Question 10===== Establish the formulas below by mathematical induction, $\left(\begin{array}{1}5 \\5 \end{array}\right)+\left(\begin{array}{l}6 \\ 5\end{array}\right)+\left(\begin{array}{l}7 \\ 5\end{array}\right)+\ldots+\left(\beg
- Question 11 Exercise 7.1
- PTBB) Peshawar, Pakistan. =====Question 11===== \begin{align} & \left(\begin{array}{l} 2 \\ 2 \end{array}\right)+\left(\begin{array}{l} 3 \\ 2 \end{array}\right)+\left(\begin{array}{l} 4 \\ 2 \end{array}\right)+\ldots+\left(
- Question 7 Exercise 7.2
- t{3})^5$ ====Solution==== Using binomial formula \begin{align}(2+\sqrt{3})^5+(2 \cdot \sqrt{3})^5& =[(... \cdot(\sqrt{3})^5\end{align} simplifing, we get \begin{align} & =2 \cdot 2^5+2^5 C_2 \cdot 2^3 \cdot(... 2})^{-}$ ====Solution==== Using binomial formula \begin{align} (1+\sqrt{2})^4-(1-\sqrt{2})^4 & =[1+{ }... \cdot(\sqrt{2})^4]\end{align} simplifing, we get \begin{align} & =2^4 C_1 \cdot \sqrt{2}+2^4 C_3 \cdot
- Question 3 Exercise 7.2
- x^2}{3}$ and $b=-\dfrac{3}{2 x}$. Let $T_{r+1}$ be the term independent of $x$ in the given expansion. $T_{r+1}$ of the given expansion is: \begin{align}T_{r+1}&=\dfrac{9 !}{(9-r) ! r !}(\dfrac... rrow r=6 $$ Putting $r=6$ in the above $T_{r+1}$ \begin{align}T_{6-1}&=\dfrac{9 !}{(9-6) ! 6 !}\cdot \... 84 \cdot \dfrac{4^7}{3^3} \cdot \dfrac{3^6}{2^6} \because(-3)^6=3^6 \\ \Rightarrow T_7&=84 \cdot \dfra
- Question 4 Exercise 7.2
- n $n=20, \quad a=x^2$ and $b=-x$. Let $T_{r, 1}$ be the term containing $x^{23}$ that is: \begin{align}T_{r-1}&=\dfrac{20 !}{(20-r) ! r !}(x^2)^{20 r}(-x... _{r-1}$ containing $x^{33}$ is possibic only if \begin{align}x^{40 \cdot r}&=x^{23}\\ \Rightarrow 40-... 7\end{align} Putting $r=17$. in $T_{r+1}$ we get \begin{align}T_{17-1}&=\dfrac{20 !}{(20 \cdot 17) ! 1
- Question 10 Exercise 7.3
- tion: The given series is binomial series. Let it be identical with the expansion of $(1+x)^n$ that is $$ \begin{aligned} & 1+n x+\frac{n(n-1)}{2 !} x^2 \\ & +... rac{1}{16}$ Dividing Eq.(2) by Eq.(3), we get $$ \begin{aligned} & \frac{n-1}{2 n}=\frac{3}{32} \cdot ... $ Putting $n=-\frac{1}{2}$ in Eq.(1), we get $$ \begin{aligned} & -\frac{1}{2} x=-\frac{1}{4} \\ & \R
- Question 5 and 6 Exercise 7.3
- 5 If $x$ is such that $x^2$ ard higher of $x$ may be negleeled. then show that $$ \frac{(8+3 x)^{\frac... 3}-3 x j^{\frac{2}{3}}}{2 \cdot 3 x+4-5 x} $$ $$ \begin{aligned} & =\frac{8^{\frac{2}{3}}\left(1+\frac... and neglecting $x^2$ and higher powers of $x$ $$ \begin{aligned} & =\left(1-\frac{3 x}{2}+\frac{x}{4}+... cting $x^2$ and higher powers of $x$, we have $$ \begin{aligned} & =1+\frac{5 x}{8}-\frac{5 x}{x} \\ &
- Question 1 Review Exercise 7
- ollapse> ii. How many two digits odd numbers can be formed form the digits $\{1,2,3,4,5,6,7\}$ if rep... </collapse> iii. How many six digits number can be formed from the digits $\{1,2,3,4,6,7,8\}$ withou... e> v. In how many different ways can $5$ couples be seated around a circular table if the couple must not be separated? * (a) $768$ * %%(b)%% $724
- Question 12 Exercise 7.3
- $ Now the above series is binomial series. Lel it be identical with the expansion of $(1+x)^n$ that is $$ \begin{aligned} & 10+n x+\frac{n(n-1)}{2 !} x^2+ \\ &... rac{1}{16}$ Dividing Eq.(2) by Eq.(3), we get $$ \begin{aligned} & \frac{n-1}{2 n}=\frac{1.3}{2 !} \cd... \frac{1}{4} \Rightarrow x=-\frac{1}{2}$. Thus $$ \begin{aligned} & 2 y+1=\left(1-\frac{1}{2}\right)^{-
- Question 12 Exercise 7.1
- \mathbb{Z}$$ Thus it is true for $n=1$ 2. Let it be true for $n=k>1$ then $$\dfrac{5^{2 k}-1}{24} \in \mathbb{Z}$$ 3. For $n=k+1$ then consider \begin{align}\dfrac{5^{2(k+1)}-1}{24}&=\dfrac{5^{2 k+... s an integer. ====Solution==== 1. For $n=1$ then \begin{align}\dfrac{10^{n+1}-9 n-10}{81}&=\dfrac{10^{... \end{align} Thus it is true for $n=1$. 2. Let it be true for $n=k$, then \begin{align}\dfrac{10^{k+1}
- Question 1 Exercise 7.3
- Using binomial theorem to tind the four terms $$ \begin{aligned} & (1-x)^{\frac{1}{2}}=1+\frac{1}{2} x... c{1}{2}-1\right)}{2 !}(-x)^2 \end{aligned} $$ $$ \begin{aligned} & +\frac{-\frac{1}{2}\left(-\frac{1}{... aligned} $$ Solution: Using binomial theorem $$ \begin{aligned} & (1-x)^{\frac{3}{2}}=1-\frac{3}{2} x... $$ (iii) $(8+12 x)^{\frac{2}{3}}$ Solution: $$ \begin{aligned} & (8+12 x)^{\frac{2}{3}}=8^{\frac{2}{
- Question 2 Exercise 7.3
- laces. (i) $\sqrt{26}$ Solution: We are given $$ \begin{aligned} & \sqrt{26}=\sqrt{25+1} \\ & =\sqrt{2... }} \end{aligned} $$ Using binomial expansion $$ \begin{aligned} & \sqrt{26}=5\left[1+\frac{1}{25}\rig... } \text {. } $$ Using binomial expansion now $$ \begin{aligned} & =1-\frac{1}{2}(-0.002)+ \\ & \frac{... ecimal places. Solution: The cube root of 126 can be written as: \begin{aligned} & =5\left[1+\frac{1}{
- Question 7 & 8 Review Exercise 7
- des 4. Hence given is true for $n=1$. (2.) Let it be true for $n=k>1$ then $7^n-3^n=4 Q$ where $Q$ is ... ion hypothesis. (3.) For $n=k+1$ then we have $$ \begin{aligned} & 7^{k+1}-3^{k+1}=7.7^k-3.3^k \\ & =(... .3^k \\ & =4.7^k+3.7^k-3.3^k \end{aligned} $$ $$ \begin{aligned} & =4.7^k+3\left[7^k-3^k\right] \\ & \... .4 Q \end{aligned} $$ by induction hypothesis $$ \begin{aligned} & \Rightarrow 7^{k=1}-3^{k+1}=4\left[
- Question 13 Exercise 7.1
- the given statement is true for $n=1$. 2. Let it be true for $n=l>I$ then $2^k>k\cdots(i)$ 3. For $n=k+1$ then we consider \begin{align} & 2^{k+1}=2^k \cdot 2>k \cdot 2 \quad \... e given proposition is true for $n=4$. 2. Let it be true for $n=k>4$ then $k !>k^2\cdots(i)$ 3. For $n=k+1$ then we have \begin{align} & (k+1) !=(k+1) k !>(k+1)(k+1) \\ & \be