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- Question 3 Exercise 7.2
- nt of $x$ is possible only if $x^{18-3 r}=x^0$ $$\Rightarrow 18-3 r=0 \Rightarrow 3 r=18 \Rightarrow r=6 $$ Putting $r=6$ in the above $T_{r+1}$ \begin{align}T_{6-1}&=\dfrac{9 !}{(9-6) ! 6 !}... ^3} \cdot \dfrac{3^6}{2^6} \because(-3)^6=3^6 \\ \Rightarrow T_7&=84 \cdot \dfrac{2^6}{2^6} \cdot 3^{6-3} \bec
- Question 10 Exercise 7.3
- n-1}{2 n}=\frac{3}{32} \cdot 16=\frac{3}{2} \\ & \Rightarrow 3 n=n-1 \Rightarrow n=-\frac{1}{2} . \end{aligned} $$ Putting $n=-\frac{1}{2}$ in Eq.(1), we get $$ \begin{aligned} & -\frac{1}{2} x=-\frac{1}{4} \\ & \Rightarrow x=\frac{1}{2} . \text { Thus } \\ & \left(1+\frac... {1 \cdot 3}{2 !} \cdot \frac{1}{2^4}+\ldots \\ & \Rightarrow\left(\frac{3}{2}\right)^{\frac{1}{2}}=1-\frac{1}{
- Question 12 Exercise 7.3
- !} \cdot \frac{1}{2^4} \cdot 16=\frac{3}{2} \\ & \Rightarrow 6 n=2 n-2 \text { or } n=-\frac{1}{2} . \end{alig... }$ in Eq.(1), we get $-\frac{1}{2} x=\frac{1}{4} \Rightarrow x=-\frac{1}{2}$. Thus $$ \begin{aligned} & 2 y+1=\left(1-\frac{1}{2}\right)^{-\frac{1}{2}} \\ & \Rightarrow 2 y+1=\left(\frac{1}{2}\right)^{-\frac{1}{2}} \\ ... !} \cdot \frac{1}{2^4} \cdot 16=\frac{3}{2} \\ & \Rightarrow 6 n=2 n-2 \text { or } n=-\frac{1}{2} . \end{alig
- Question 4 Exercise 7.2
- c only if \begin{align}x^{40 \cdot r}&=x^{23}\\ \Rightarrow 40-r&=23\\ \Rightarrow r&=40-23=17\end{align} Putting $r=17$. in $T_{r+1}$ we get \begin{align}T_{17-1}&=\dfr... 0 !}{(20 \cdot 17) ! 17 !}(-1)^{17} x^{40-17} \\ \Rightarrow T_{18}&=-1140 x^{23}\end{align} Hence the coeffic... f \begin{align}\dfrac{1}{x^r}&=\dfrac{1}{x^4}\\ \Rightarrow 4=r\end{align} Putting $r=4$ in $T_{r+ 1}$ we get
- Question 2 Exercise 7.2
- _{3+1}=\dfrac{7 !}{(7-3) ! 3 !} 2^{7-3} a^3 \\ & \Rightarrow T_4=\dfrac{7 !}{4 ! 3 !} \cdot 2^4 a^3 \\ & \Rightarrow T_4=35 \times 16 a^3 \\ & \Rightarrow T_4=560 a ^3 \end{align} =====Question 2(ii)===== Find the indicate te... t of $x$ is possible conly if $x^{21-3 t}=x^0$ $$\Rightarrow 21-3 r=0 \Rightarrow r=7 $$ Putting $r=7$ in the
- Question 3 & 4 Review Exercise 7
- t\left(2^4\right) \cdot(-4)^3 \cdot x^4 y^3 \\ & \Rightarrow T_4=-35840 x^4 y^3 . \end{aligned} $$ Q4 $2^7 x ... only if $$ x^{4 \prime \prime} y^{\prime}=x y^3 \Rightarrow r=3 \text {. } $$ Thus the term is: $$ \begin{al... 3 !} 2^3 a^4 x^{4-3} y^3 . \\ & =2^7 x y^3 \\ & \Rightarrow 4 \cdot 2^3 \cdot a^4=2^7 \\ & \Rightarrow a^4=\frac{2^7}{2^3 \cdot 2^2}=2^2 \\ & \Rightarrow a^2=\sqrt{2^2}=
- Question 14 Exercise 7.3
- } \\ & -q(1+q h+\text { higher powcrs } h) \\ & \Rightarrow p x^p-q x^q=p+p^2 h-q-q^2 h \\ & \Rightarrow p x^p-q x^q=(p-q)+\left(p^2-q^2\right) h \\ & \Rightarrow p x^p-q x^q \\ & -(p-q)+[(p-q)(p+q)] h \\ & \Rightarrow p x^p-q x^q=(p-q)[1+(p+q) h] \\ & \Rightarrow p x^p-q
- Question 11 Exercise 7.3
- !} \cdot \frac{1}{2^4} \cdot 16=\frac{3}{2} \\ & \Rightarrow 6 n=2 n-2 \text { or } n=-\frac{1}{2} . \\ & \tex... Eq.(1), we get } \\ & -\frac{1}{2} x=\frac{1}{4} \Rightarrow x=-\frac{1}{2} \text {. Thus } \\ & y+1=\left(1-\... e both sides $$ \begin{aligned} & (y+1)^2=2 \\ & \Rightarrow y^2+2 y+1-2=0 \\ & \Rightarrow y^2+2 y-1=0 . \end{aligned} $$ Which is the desired result. ====Go To====
- Question 7 & 8 Review Exercise 7
- gin{aligned} & =4.7^k+3\left[7^k-3^k\right] \\ & \Rightarrow 7^{k+1}-3^{k+1}=4.7^k+3.4 Q \end{aligned} $$ by induction hypothesis $$ \begin{aligned} & \Rightarrow 7^{k=1}-3^{k+1}=4\left[7^k+3 Q\right] \\ & \Rightarrow 7^{k-1}-3^{k+1} \text { is divisible by } 4 . \end{ali... 2 \\ & =1+(k+1) x+k x^2 \\ & \geq 1+(k+1) x \\ & \Rightarrow(1+x)^{k+1} \geq[1+(k+1) x] . \end{aligned} $$ He
- Question 13 Exercise 7.1
- ^k \cdot 2>k \cdot 2 \quad \text { by (i) } \\ & \Rightarrow 2^{k+1}>2 k=k+k \\ &\Rightarrow 2^{k+1}>k+1 \text {. as } k>1\end{align} Which is the form of proposition tak... =(k+1) k !>(k+1)(k+1) \\ & \because k !>k+1 \\ & \Rightarrow(k+1) !>(k+1)^2 . \end{align} Which is the form ta
- Question 5 & 6 Review Exercise 7
- t of $x$ is possible only if $x^{4 r-20}=x^0$ $$ \Rightarrow 4 r-20=0 \Rightarrow r=5 \text {. } $$ Putting in $T_{r+1}$ we get $$ T_{5+1}=\frac{10 !}{5 ! 5 !} \cdot 2^0 \cdot x^0 $$ $$ \Rightarrow \quad T_6=252 . $$ Hence $T_6$ is constant term
- Question 15 Exercise 7.1
- 1$, then $$a^{2 n}-b^{2 m}=a^2-b^2=(a+b)(a-b)$$ $\Rightarrow(a+b)$ is a factor of $a^2-b^2$. Thus it is true ... induction hypothesis $$=(a+b)[a^2 Q-b^2(a-b)]$$ $\Rightarrow a+b$ is factor of $a^{2 n}-b^{2 m}$ for $m=k+1$.
- Question 6 Exercise 7.2
- ly is \begin{align}x^{\dfrac{23-4 r}{2}}&=x^0\\ \Rightarrow 23-4 r&=0 \\ \Rightarrow r&=\dfrac{23}{4}\end{align} Which is not possible because $r$ should be a positive i
- Question 2 Exercise 7.1
- k+1 \\ & =2 k^2+2 k+k+1 \\ & =2 k(k+1)+1(k+1) \\ \Rightarrow 1+5+9+\ldots+(4 k-3)+(4 k+1)& =(k+1)[2 k+1] \\ &
- Question 3 Exercise 7.1
- [\dfrac{k}{2}+1] \\ & =3(k+1)[\dfrac{k+2}{2}] \\ \Rightarrow 3+6+9+\ldots+3 k+3(k+1) & =\dfrac{3(k+1)(k+1-1)}{