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- Question 4 Exercise 7.2
- $ be the term containing $x^{23}$ that is: \begin{align}T_{r-1}&=\dfrac{20 !}{(20-r) ! r !}(x^2)^{20 r}(-... & =\dfrac{20 !}{(20-r) ! r !}(-1)^r x^{40-r}\end{align} But $T_{r-1}$ containing $x^{33}$ is possibic only if \begin{align}x^{40 \cdot r}&=x^{23}\\ \Rightarrow 40-r&=23\\ \Rightarrow r&=40-23=17\end{align} Putting $r=17$. in $T_{r+1}$ we get \begin{align
- Question 3 Exercise 7.2
- ion. $T_{r+1}$ of the given expansion is: \begin{align}T_{r+1}&=\dfrac{9 !}{(9-r) ! r !}(\dfrac{4 x^2}{3... 3^{9-r}} \cdot \dfrac{(-3)^r}{2^r} x^{18-3 r}\end{align} But the term $T_{r+1}$ independent of $x$ is pos... =6 $$ Putting $r=6$ in the above $T_{r+1}$ \begin{align}T_{6-1}&=\dfrac{9 !}{(9-6) ! 6 !}\cdot \dfrac{4^{... (2^2)^3 \\ \Rightarrow T_7&=84 \times 27=2268\end{align} Hence $T_7$ is independent of $x$ and is $2268.$
- Question 12 Exercise 7.1
- \mathbb{Z}$$ 3. For $n=k+1$ then consider \begin{align}\dfrac{5^{2(k+1)}-1}{24}&=\dfrac{5^{2 k+2}-1}{24}... k}-1}{24} \\ & =5^{2 k}+\dfrac{5^{2 k}-1}{24}\end{align} Clearly $5^{2 k} \in \mathbb{Z} \quad \forall k ... nteger. ====Solution==== 1. For $n=1$ then \begin{align}\dfrac{10^{n+1}-9 n-10}{81}&=\dfrac{10^{i+1}-9.1-... } \\ & =\dfrac{100-9-10}{81}=1 \in \mathbb{Z}\end{align} Thus it is true for $n=1$. 2. Let it be true fo
- Question 2 Exercise 7.2
- For $4^{\text {th }}$ term, putting $r=3$ \begin{align} & T_{3+1}=\dfrac{7 !}{(7-3) ! 3 !} 2^{7-3} a^3 \... \times 16 a^3 \\ & \Rightarrow T_4=560 a ^3 \end{align} =====Question 2(ii)===== Find the indicate term... For $8^{\text {th }}$ term, putting $r=7$ \begin{align}T_{7+1}&=\dfrac{10 !}{(10-7) ! 7 !}(\dfrac{x}{2})... }^{10}C_7 \,2^{-3}\cdot 3^7 x^3 \cdot y^{-7} \end{align} =====Question 2(iii)===== Find the indicate t
- Question 5 Exercise 7.2
- r$$ To get middle term $T_5$, we put $r=4$ \begin{align}T_5&=\dfrac{8 !}{(8-4) ! 4 !}(\dfrac{a}{x})^{8-4}... 4}{x^4} \cdot b^4 x^4 \\ & =70 \cdot a^4 b^4 \end{align} Thus $T_5$ is the middle term of the expansion w... r$$ Putting $r=4$ to get first middle term \begin{align} T_5&=\dfrac{9 !}{(9-4) ! 4 !}(3 x)^{9-4}(-\dfrac... 8}{16} x^{13}\\ & T_5=\dfrac{15309}{8} x^{13}\end{align} Putting $r=5$ to get the $2^{\text {nd }}$ middl
- Question 7 Exercise 7.2
- 5$ ====Solution==== Using binomial formula \begin{align}(2+\sqrt{3})^5+(2 \cdot \sqrt{3})^5& =[(2)^5+{ }^... cdot(\sqrt{3})^4+2{ }^5 C_5 \cdot(\sqrt{3})^5\end{align} simplifing, we get \begin{align} & =2 \cdot 2^5+2^5 C_2 \cdot 2^3 \cdot(\sqrt{3})^2+2^5 C_4 \cdot 2 \cd... ext { Thus }(2+\sqrt{3})^5+(2-\sqrt{3})^5=724\end{align} =====Question 7(ii)===== $(1+\sqrt{2})^4-(1-\sq
- Question 10 Exercise 7.1
- right)$ ====Solution==== 1. For $n=1$ then \begin{align}\left(\begin{array}{l}5 \\ 5\end{array}\right)&=\... end{array}\right)&=\dfrac{6 !}{(6-6) ! 6 !}=1\end{align} Thus it is true for $n=1$. 2. Let it be true for $n=k$ then \begin{align}\left(\begin{array}{l}5 \\ 5\end{array}\right)+\l... {array}{c}k+5 \\ 6\end{array}\right) . ...(i)\end{align} 3. For $n=k+1$ then $(k+1)^{t h}$ term of the se
- Question 11 Exercise 7.1
- Peshawar, Pakistan. =====Question 11===== \begin{align} & \left(\begin{array}{l} 2 \\ 2 \end{array}\righ... n+1 \\ 3 \end{array}\right), \quad n \geq 2 \end{align} ====Solution==== 1. For $n=2$ then \begin{align} \left(\begin{array}{l} 2 \\ 2 \end{array}\right)&=\dfr... end{array}\right)&=\dfrac{3 !}{(3-3) ! 3 !}=1\end{align} Thus it is true for $n=1$. 2. Let it be true fo
- Question 1 Exercise 7.2
- 4$ ====Solution==== Using binomial theorem \begin{align}(x^2-\dfrac{1}{y})^4&=(x^2)^4+{ }^4 C_1(x^2)^3(-\... 6x^4}{y^2}- \dfrac{4x^2}{y^3}+\dfrac{1}{y^4} \end{align} =====Question 1(ii)===== Expand by using Binomi... 7$ ====Solution==== Using binonial theorem \begin{align} & (1+x y)^7=1+{ }^7 C_1(1)^6(x y)+{ }^7 C_2 \cdo... y^4 +21 x^5 y^5+7 x^6 y^6+x^7 y^7 \text {. } \end{align} =====Question 1(iii)===== Expand by using Binom
- Question 2 Exercise 7.1
- $n=1$. 2. Let it be true for $n=k$, then \begin{align}1+5+9+\ldots+(4 k-3)\\ & =k(2 k-1)....(i) \\ \end{align} 3. For $n=k+1$, the $k+1$ term of the series, wh... both sides of the induction hypcthesis (i) \begin{align}1+5+9+\ldots+(4 k-3)+(4 k+1)& =k(2 k-1)+4 k+1 \\ ... +(4 k+1)& =(k+1)[2 k+1] \\ & =(k+1)[2(k+1)-1]\end{align} Which is the form taken by proposition when $n$
- Question 4 Exercise 7.1
- =1$. 2. Let it be true for $n=k$, we have \begin{align}3+7+11+\cdots+(4 k-1) & =k(2 k+1)....(i) \end{align} 3. Now considering for $n=k+1$, the $(k+1)$ term o... s of the induction hypothesis (i), we have \begin{align} 3+7+11+\cdots+(4 k-1)+[4(k+1)-1] & =k(2 k+1)+4(k... 1+\cdots+(4 k-1)+[4(k+ 1)-1]&=(k+1)[2(k+1)+1]\end{align} Which is the form of given statement when $n$ is
- Question 6 Exercise 7.1
- 2. Let it be true for $n=k$, then we have \begin{align}1(1 !)+2(2 !)+3(3 !)+\ldots+k(k !)& =(k+1) !-1 \ldots . .(i)\end{align} 3. For $n=k+1$ the $(k+1)^{t h}$ term of the ser... es of the induction hypothesis(i), we have \begin{align}1(1 !)+2(2 !)+3(3 !)+\ldots+k(k !)+ (k+1) !(k+1) ... k !)+(k+1)[(k+1) !] & =(k+2) !-1=(k+1+1) !-1 \end{align} Which is the form proposition when $n$ is replac
- Question 7 Exercise 7.1
- $n=1$. 2. Let it be true for $n=k$, then \begin{align}1.2+2.3+3.4+\ldots+k(k+1)& =\dfrac{k(k+1)(k+2)}{3}....(i)\end{align} 3. Considering for $n=k+1$, then $(k-1)^{t h}$ t... s of the induction hypothesis (i), we have \begin{align}1.2+2.3+3.4+\ldots+k(k+1)+(k+1)(k+2) & =\dfrac{k(... )+(k+1)(k+2)& =\dfrac{(k+1)(k+1+1)(k+1+2)}{3}\end{align} Which is the form of the proposition when $n$ is
- Question 8 Exercise 7.1
- $n=1$. 2. Let it be true for $n-k>1$ then \begin{align}1+2+2^2+2^3+\ldots+2^{k-1} \\ & =2^k-1 ....(i)\end{align} 3. Considering for $n-k-1$, then $(k+1)^{t h}$ t... s of the induction hypothesis (i), we have \begin{align}1+2+2^2+2^3+\ldots+2^{k-1}-2^k & =2^k-12^k \\ & =... & 1+2+2^2+2^3+\ldots+2^{k-1}+2^k & =2^{k+1}-1\end{align} Which is the form of the proposition when $n$ is
- Question 13 Exercise 7.1
- cdots(i)$ 3. For $n=k+1$ then we consider \begin{align} & 2^{k+1}=2^k \cdot 2>k \cdot 2 \quad \text { by... \\ &\Rightarrow 2^{k+1}>k+1 \text {. as } k>1\end{align} Which is the form of proposition taken when $n$ ... k^2\cdots(i)$ 3. For $n=k+1$ then we have \begin{align} & (k+1) !=(k+1) k !>(k+1)(k+1) \\ & \because k !>k+1 \\ & \Rightarrow(k+1) !>(k+1)^2 . \end{align} Which is the form taken by proposition when $n$