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- Question 9 Exercise 7.1
- h the formulas below by mathematical induction, $\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ldots+\dfrac{1}{3^n}=\dfrac{1}{2}[1-\dfrac{1}{3^n}]$ ====Solution==== 1. For $n=1$ then $$\dfrac
- Question 1 Exercise 7.2
- (i)===== Expand by using Binomial theorem: $(x^2-\dfrac{1}{y})^4$ ====Solution==== Using binomial theorem \begin{align}(x^2-\dfrac{1}{y})^4&=(x^2)^4+{ }^4 C_1(x^2)^3(-\dfrac{1}{y})+ \\ & { }^4 C_2(x^2)^2(-\dfrac{1}{y})^2+{ }^4 C_3(x^2)(-\dfrac{1}{y})^3 + { }^4 C_4(-\dfrac{1}{y}
- Question 3 Exercise 7.2
- d the term independent of $x$ in the expansion $(\dfrac{4 x^2}{3}-\dfrac{3}{2 x})$ ====Solution==== In the above expansion $n=9, \quad a=\dfrac{4 x^2}{3}$ and $b=-\dfrac{3}{2 x}$. Let $T_{r+1}$ be the term independent of $x$ in the given expansion
- Question 5 Exercise 7.2
- 5(i)===== Find middle term in the expansion of $(\dfrac{a}{x}+b x)^8$ ====Solution==== Since we see that $a=\dfrac{a}{x}$. $b=b x$ and $n=8$ Since $n-8$ is a the ... n are $8+1=9$ The middle term is only one so, $$(\dfrac{8+2}{2})^{t h}=5^{t h}$. Now $T_{r+1}$ of the given expansion is: $$T_{r+1}=\dfrac{8 !}{(8-r) ! r !}(\dfrac{a}{x})^{8-r}(b x)^r$$ To
- Question 12 Exercise 7.1
- 12(i)===== Show by mathematical induction that $\dfrac{5^{2 n}-1}{24}$ is an integer. Solution: 1. For $n=1$, then $$\dfrac{5^{2 n}-1}{24}=\dfrac{5^{2.1}-1}{24}=\dfrac{24}{24}=1 \in \mathbb{Z}$$ Thus it is true for $n=1$ 2. Let it be true for $n=k>1
- Question 4 Exercise 7.2
- ntaining $x^{23}$ that is: \begin{align}T_{r-1}&=\dfrac{20 !}{(20-r) ! r !}(x^2)^{20 r}(-x)^r \\ & =\dfrac{20 !}{(20-r) ! r !}(-1)^r \cdot x^{40-2 r+r} \\ & =\dfrac{20 !}{(20-r) ! r !}(-1)^r x^{40-r}\end{align} But... =17$. in $T_{r+1}$ we get \begin{align}T_{17-1}&=\dfrac{20 !}{(20 \cdot 17) ! 17 !}(-1)^{17} x^{40-17} \\
- Question 2 Exercise 7.2
- }}$ term, putting $r=3$ \begin{align} & T_{3+1}=\dfrac{7 !}{(7-3) ! 3 !} 2^{7-3} a^3 \\ & \Rightarrow T_4=\dfrac{7 !}{4 ! 3 !} \cdot 2^4 a^3 \\ & \Rightarrow T_4=... erm in the expansion $8^{\text {th }}$ term in $(\dfrac{x}{2}-\dfrac{3}{y})^{10}$ ====Solution==== In the above $n=10$, $a=\dfrac{x}{2}$ and $b=-\dfrac{3}{y}$.
- Question 10 Exercise 7.1
- }\left(\begin{array}{l}5 \\ 5\end{array}\right)&=\dfrac{5 !}{(5-5) ! 5 !}=1\\ \left(\begin{array}{c}1+5 \\ 6\end{array}\right)&=\dfrac{6 !}{(6-6) ! 6 !}=1\end{align} Thus it is true fo... \begin{array}{c}k+5 \\5 \end{array}\right) \\ & =\dfrac{(k+5) !}{(k+5-6) ! 6 !}+\dfrac{(k+5) !}{(k+5-5) ! 5 !} \\ & =\dfrac{(k+5) !}{(k-1) ! 6.5 !}+\dfrac{(k+5)
- Question 6 Exercise 7.2
- e constant term in the expansion or $(2 \sqrt{x}-\dfrac{3}{x \sqrt{x}})^{23}$ ====Solution==== Since we see that $a=2 \sqrt{x}$, $b=-\dfrac{3}{x \sqrt{x}}$ and $n=23$. By constant term, we... f the given expansion is: \begin{align} T_{r+1}&=\dfrac{23 !}{(23-r) ! r !}(2 \sqrt{x})^{23-r}(-\dfrac{3}{x \sqrt{x}})^r \\ & =\dfrac{23 !}{(23-r) ! r !} \cdot
- Question 11 Exercise 7.1
- left(\begin{array}{l} 2 \\ 2 \end{array}\right)&=\dfrac{2 !}{(2-2) ! 2 !}=1 \\ \left(\begin{array}{c} 2+1 \\ 2 \end{array}\right)&=\dfrac{3 !}{(3-3) ! 3 !}=1\end{align} Thus it is true fo... egin{array}{c} k+1 \\ 2 \end{array}\right) \\ & =\dfrac{(k+1) !}{(k+1-3) ! 3 !}+\dfrac{(k+1) !}{(k+1-2) ! 2 !} \\ & =\dfrac{(k+1) !}{(k-2) ! 3 !}+\dfrac{(k+1) !
- Question 5 Exercise 7.1
- matical induction, $1^3+2^3+3^3+\ldots+n^3=\left[\dfrac{n(n+1)}{2}\right]^2$. ====Solution==== 1. For $n=1$, then $1^3=1=\left[\dfrac{1(1+1)}{2}\right]^2=1$. Thus it is true for $n=1... _1$, then \begin{align}1^3+2^3+3^3+\ldots+k^3& =[\dfrac{k(k+1)}{2}]^2....(i)\end{aligned} 3. Now $n=k+1$ ... \begin{align}1^3+2^3+3^3+\ldots+k^3+(k+1)^3 & =[\dfrac{k(k+1)}{2}]^2+(k+1)^3 \\ & =(k+1)^2[\dfrac{k^2}{4
- Question 3 Exercise 7.1
- elow by mathematical induction $3+6+9+\ldots+3 n=\dfrac{3 n(n+1)}{2}$ ====Solution==== 1. For $n=1$ then $3=\dfrac{3.1(1+1)}{2}=3$, thus the above statement or pro... it be true for $n=k$, we have $$3+6+9+\ldots+3 k=\dfrac{3 k(k+1)}{2}....(i)$$ 3. Now considering for $n=k... we have \begin{align}3+6+9+\ldots+3 k+3(k+1) & =\dfrac{3 k(k+1)}{2}+3(k+1) \\ & =3(k+1)[\dfrac{k}{2}+1]
- Question 7 Exercise 7.1
- thematical induction, $1.2+2.3+3.4+\ldots+n(n+1)=\dfrac{n(n+1)(n+2)}{3}$ ====Solution==== 1. For $n=1$ then $$1.2=2=\dfrac{1(1+1)(1+2)}{3}=2 $$ Thus it is true for $n=1$. ... $, 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 ... in{align}1.2+2.3+3.4+\ldots+k(k+1)+(k+1)(k+2) & =\dfrac{k(k+1)(k-2)}{3}+(k+1)(k+2) \\ & =(k-1)(k+2)[\dfra
- Question 1 Review Exercise 7
- psed="true">(a): $120$</collapse> iv. Evaluate $\dfrac{(n+2) !(n-2) !}{(n+1) !(n-1) !}$ * (a) $(n-3)$ * %%(b)%% $(\dot{n}-1)$ * %%(c)%% $\dfrac{n+1}{n+2}$ * (d) $\dfrac{n+2}{n-1}$ \\ <btn type="link" collapse="a4">See Answer</btn><collapse id="a4" collapsed="true">(d): $\dfrac{n+2}{n-1}$ </collapse> v. In how many different