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- Question 1 Review Exercise 7
- correct option. <panel> i. In how many ways can we name the vertices of pentagon using any five of t... ) $4880$ \\ <btn type="link" collapse="a1">See Answer</btn><collapse id="a1" collapsed="true">(a): $25... ts $\{1,2,3,4,5,6,7\}$ if repeated digits are allowed? * %%(a)%% $14$ * (b) $42$ * %%(c)... (d) $21$ \\ <btn type="link" collapse="a2">See Answer</btn><collapse id="a2" collapsed="true">(c): $28
- Question 5 and 6 Exercise 7.3
- omial expansion and neglecting $x^2$ and higher powers of $x$. $-\left(1+\frac{2}{3} \cdot \frac{3 x}{8}+\right.$ higher powers of $\left.x\right) x$ $\left(1-\frac{3 x}{2}+\right.$ higher powers of $\left.x\right) \times$ $\left(1+\frac{1}{2}, \frac{5 x}{4}+\right.$ higher powers of $\left.x\right)$ $-\left(1+\frac{x}{4}\right
- Question 12 Exercise 7.3
- \ldots$ then show that $4 y^2+4 y-1=0$. Solution: We are given $$ 2 y=\frac{1}{2^2}+\frac{1.3}{2 !} \c... $$ Adding 1 to both sides of the above equation, we get $S=2 y+1=1+\frac{1}{2^2}+\frac{1.3}{2 !} \cdo... dots \end{aligned} $$ Comparing both the series, we have $n x=\frac{1}{2^2}=\frac{1}{4}.... (1)$ and ... \cdot \frac{1}{2^4} $$ Taking square of Eq.(1), we have $n^2 x^2=\frac{1}{16}$ Dividing Eq.(2) by Eq
- Question 10 Exercise 7.3
- dots \end{aligned} $$ Comparing both the series, we have $n x=-\frac{1}{4}$ (I) and $\frac{n(n-1)}{2 ... !} \cdot \frac{1}{2^4}$ Taking square of Eq.(1), we have $n^2 x^2=\frac{1}{16}$ Dividing Eq.(2) by Eq.(3), we get $$ \begin{aligned} & \frac{n-1}{2 n}=\frac{3}... {aligned} $$ Putting $n=-\frac{1}{2}$ in Eq.(1), we get $$ \begin{aligned} & -\frac{1}{2} x=-\frac{1}
- Question 10 Exercise 7.2
- f even binomial cosficient $s=2^{n-1}$. Solution: We know that $$ \left.(1+x)^n=\left(\begin{array}{l}... ^n \cdot $$ Putting $x=1$ in the above equation, we have $(1 \div 1)^n=\left(\begin{array}{l}n \\ 0\e... ows that the sum of the :nefficiens is $?^n$. Now we know that $$ \begin{aligned} & (1+x)^n=\left(\beg... \end{array}\right) 1^n \\ & \end{aligned} $$ If we put $x=-1$ in the above eyuation, we get $$ \begi
- Question 7 and 8 Exercise 7.3
- BB) Peshawar, Pakistan. Q7 II $x^4$ and higher powers are neglected and $(1-x)^{\frac{1}{4}}+(1-x)^{\... 1}{4}}=a-b x^2$ then find $a$, and $b$. Solution: We are taking L.H.S of the above given equation and ... ned} $$ Adding and neglecting $x^4$ and higher powers, we get $-2-\frac{3 x^2}{16}$ Hence $(1-x) \frac{1}{4}-(1-x) \frac{1}{4}=a-b x^2$ and tve oltain: $
- Question 5 Exercise 7.2
- of $(\dfrac{a}{x}+b x)^8$ ====Solution==== Since we see that $a=\dfrac{a}{x}$. $b=b x$ and $n=8$ Si... c{a}{x})^{8-r}(b x)^r$$ To get middle term $T_5$, we put $r=4$ \begin{align}T_5&=\dfrac{8 !}{(8-4) ! 4... f $(3 x-\dfrac{x^2}{2})^9$ ====Solution==== Since we see that $a=3 x$, $b=-\dfrac{x^2}{2}$ and $n=9$. ... terms in the expansion are $9+1=10$. So in this we have two middle terms that are $(\dfrac{9+1}{2})^
- Question 11 Exercise 7.3
- }+\ldots$ then show that $y^2+2 y-1=0$. Solution: We are given $y=\frac{1}{2^2}+\frac{1.3}{2 !} \cdot ... -2))}{3 !} x^3+\ldots$ Comparing both the series, we have $n x=\frac{1}{2^2}=\frac{1}{4}$ (1) and $\fr... !} \cdot \frac{1}{2^4}$ Taking square of Eq.(1), we have $n^2 x^2=\frac{1}{16}$ Dividing Eq.(2) by Eq.(3), we get $$ \begin{aligned} & \frac{n-1}{2 n}=\frac{1
- Question 7 & 8 Review Exercise 7
- rove that $7^n-3^n$ is divisible by 4 . Solution: We using mathematical induction to prove the given s... n the induction hypothesis. (3.) For $n=k+1$ then we have $$ \begin{aligned} & 7^{k+1}-3^{k+1}=7.7^k-3... all natural number $n$ where $x>-1$. - Solution: We try to prove this using mathernatical induction. ... eq(1+k x) $$ 3. Now for $\boldsymbol{n}=k+1$ then we have $$ \begin{aligned} & (1+x)^{k+1}=(1+x)^k(1+x
- Question 4 Exercise 7.2
- =40-23=17\end{align} Putting $r=17$. in $T_{r+1}$ we get \begin{align}T_{17-1}&=\dfrac{20 !}{(20 \cdot... tarrow 4=r\end{align} Putting $r=4$ in $T_{r+ 1}$ we get \begin{align}T_{4+1}&=\dfrac{8 !}{(8-4)!4!}2^... xt{or} r=3\end{align} Putting $r=3$ in the above, we get \begin{align}T_{3+1}&=\dfrac{9 !}{(9-3) ! 3 !
- Question 7 Exercise 7.2
- }^5 C_5 \cdot(\sqrt{3})^5\end{align} simplifing, we get \begin{align} & =2 \cdot 2^5+2^5 C_2 \cdot 2^... }^4 C_4 \cdot(\sqrt{2})^4]\end{align} simplifing, we get \begin{align} & =2^4 C_1 \cdot \sqrt{2}+2^4 C... \right) b^5\right] \end{aligned} $$ Simplifying, we get $$ \begin{aligned} & \left.=2\left(\begin{arr
- Question 8 Exercise 7.2
- n $(3-2 x)^{10}$. when $x=\frac{3}{4}$. Solution: We first write in form $\left(3-2,1^{10}=3^{10}\left... gned} Hence by perpe:ts of like binomial theorem, we hisu that: $p+1$ - 5.. 1 - 6 icrm is numerically ... 10 \\ 5\end{array}\right) 3^{10} 5-2 \gamma^{15}$ we compute at $x=\frac{3}{4}$, so $$ \begin{aligned}
- Question 5 & 6 Review Exercise 7
- ightarrow r=5 \text {. } $$ Putting in $T_{r+1}$ we get $$ T_{5+1}=\frac{10 !}{5 ! 5 !} \cdot 2^0 \cd... the first three terms of its expansion. Solution: We can write $$ (0.99)^5=(1-0.01)^5 $$ Using binomial theorem, we have $$ \begin{aligned} & (1-0.01)^5 \cong{ }^5 C
- Question 3 Exercise 7.1
- is true for $n=1$. 2. Let it be true for $n=k$, we have $$3+6+9+\ldots+3 k=\dfrac{3 k(k+1)}{2}....(i... rm to both sides of the induction hypothesis (i), we have \begin{align}3+6+9+\ldots+3 k+3(k+1) & =\dfr
- Question 4 Exercise 7.1
- is true for $n=1$. 2. Let it be true for $n=k$, we have \begin{align}3+7+11+\cdots+(4 k-1) & =k(2 k... rm to both sides of the induction hypothesis (i), we have \begin{align} 3+7+11+\cdots+(4 k-1)+[4(k+1)-