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- Question 10 Exercise 7.2
- \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+i_n^*\right) x^n \cdot $$ ... e have $(1 \div 1)^n=\left(\begin{array}{l}n \\ 0\end{array}\right)+\left(\begin{array}{l}n \\ 1\end{ar
- Question 11 Exercise 7.1
- == \begin{align} & \left(\begin{array}{l} 2 \\ 2 \end{array}\right)+\left(\begin{array}{l} 3 \\ 2 \end{array}\right)+\left(\begin{array}{l} 4 \\ 2 \end{array}\right)+\ldots+\left(\begin{array}{l} n \\ 2 \end{array}\right)=\left(\begin{array}{c} n+1 \\ 3 \en
- Question 10 Exercise 7.1
- ematical induction, $\left(\begin{array}{1}5 \\5 \end{array}\right)+\left(\begin{array}{l}6 \\ 5\end{array}\right)+\left(\begin{array}{l}7 \\ 5\end{array}\right)+\ldots+\left(\begin{array}{c}n+4 \\ 5\end{array}\right)=\left(\begin{array}{c}n+5 \\ 6\end{
- Question 7 Exercise 7.2
- ^2 \cdot(\sqrt{3})^4+2{ }^5 C_5 \cdot(\sqrt{3})^5\end{align} simplifing, we get \begin{align} & =2 \cdo... & \text { Thus }(2+\sqrt{3})^5+(2-\sqrt{3})^5=724\end{align} =====Question 7(ii)===== $(1+\sqrt{2})^4-... 3 \cdot(\sqrt{2})^3 +{ }^4 C_4 \cdot(\sqrt{2})^4]\end{align} simplifing, we get \begin{align} & =2^4 C_... s } \\ (1+\sqrt{2})^4-(1-\sqrt{2})^4&=24 \sqrt{2}\end{align} =====Question 7(iii)===== Find $(a+b)^5+(
- Question 4 Exercise 7.2
- } \\ & =\dfrac{20 !}{(20-r) ! r !}(-1)^r x^{40-r}\end{align} But $T_{r-1}$ containing $x^{33}$ is possi... \\ \Rightarrow 40-r&=23\\ \Rightarrow r&=40-23=17\end{align} Putting $r=17$. in $T_{r+1}$ we get \begin... 17} x^{40-17} \\ \Rightarrow T_{18}&=-1140 x^{23}\end{align} Hence the coefficient of $x^{23}$ in the e... rac{8 !}{(8-r) ! r!}2^{8-r}(-1)^r(\dfrac{1}{x})^r\end{align} But $T_{r+1}$ conlaining $x^4$ is possible
- Question 10 Exercise 7.3
- !} x^2 \\ & +\frac{n(n-1(n-2))}{3 !} x^3+\ldots \end{aligned} $$ Comparing both the series, we have $... \Rightarrow 3 n=n-1 \Rightarrow n=-\frac{1}{2} . \end{aligned} $$ Putting $n=-\frac{1}{2}$ in Eq.(1), ... \frac{1 \cdot 3}{2 !} \cdot \frac{1}{2^4}+\ldots \end{aligned} $$ Hence the sum of the series is $\sqr... !} x^2 \\ & +\frac{n(n-1(n-2))}{3 !} x^3+\ldots \end{aligned} $$ Comparing both the series, we have $$
- Question 3 Exercise 7.2
- r}}{3^{9-r}} \cdot \dfrac{(-3)^r}{2^r} x^{18-3 r}\end{align} But the term $T_{r+1}$ independent of $x$ ... 4^3=(2^2)^3 \\ \Rightarrow T_7&=84 \times 27=2268\end{align} Hence $T_7$ is independent of $x$ and is $... \dfrac{10 !}{(10-r) ! r !}(-3)^r \cdot x^{10-5 r}\end{align} But the term $T_{r+1}$ independent of $x$ ... dot x^{10} 10 \\ \Rightarrow T_3&=45 \times 9=405\end{align} Hence $T_3$ is independemt of $x$ which is
- Question 1 Exercise 7.3
- rac{1}{2}\left(-\frac{1}{2}-1\right)}{2 !}(-x)^2 \end{aligned} $$ $$ \begin{aligned} & +\frac{-\frac{1}... } x+\frac{3}{8} x^2+\frac{1}{16} x^3+\ldots \\ & \end{aligned} $$ Solution: Using binomial theorem $$ ... ht)\left(\frac{3}{2}-2\right)}{3 !}(-x)^3+\ldots \end{aligned} $$ (iii) $(8+12 x)^{\frac{2}{3}}$ Solut... {3}}=4\left(1+\frac{3 x}{2}\right)^{\frac{2}{3}} \end{aligned} $$ Now using binomal expansion $$ \begi
- Question 5 and 6 Exercise 7.3
- eft(1-\frac{5 x}{4}\right)^{-\frac{1}{2}}\right] \end{aligned} $$ Applying binomial expansion and negl... \frac{5 x}{4}\right)\left(1+\frac{5 x}{8}\right) \end{aligned} $$ Multiplying and neglecting $x^2$ and... 10 x}{8} \\ & =1-\frac{5 x}{8} . \text { Hence } \end{aligned} $$ $$ \frac{(8+3 x)^{\frac{2}{3}}}{(2+3 ... ht)^{\frac{1}{2}}\left(1+\frac{1}{x}\right)^{-2} \end{aligned} $$ Applying binomial theorem and neglec
- Question 12 Exercise 7.3
- !} x^2+ \\ & \frac{n(n-1(n-2))}{3 !} x^3+\ldots \end{aligned} $$ Comparing both the series, we have $... ghtarrow 6 n=2 n-2 \text { or } n=-\frac{1}{2} . \end{aligned} $$ Putting $n=-\frac{1}{2}$ in Eq.(1), ... \ & =\left(2^{-1}\right)^{-\frac{1}{2}}-\sqrt{2} \end{aligned} $$ $$ \frac{n(n-1)}{2 !} x^2=\frac{1.3}{... ghtarrow 6 n=2 n-2 \text { or } n=-\frac{1}{2} . \end{aligned} $$ Putting $n=-\frac{1}{2}$ in Eq.(1),
- Question 2 Exercise 7.3
- {25}}=5\left[1+\frac{1}{25}\right]^{\frac{1}{2}} \end{aligned} $$ Using binomial expansion $$ \begin{a... \ & -5[1.019808 . .] \cong 5.099 \text {. } \\ & \end{aligned} $$ (ii) $\frac{1}{\sqrt{0.998}}$ Soluti... 0015+ \\ & 0.000000005+\ldots \\ & \cong 1.001 . \end{aligned} $$ (iii) Find cube root of 126 correct t... \cdot \frac{1}{2} \cdot \frac{1}{15625}+\right. \end{aligned} \begin{aligned} & \left.\frac{1}{3} \cdo
- Question 12 Exercise 7.1
- ^{2 k}-1}{24} \\ & =5^{2 k}+\dfrac{5^{2 k}-1}{24}\end{align} Clearly $5^{2 k} \in \mathbb{Z} \quad \for... }{81} \\ & =\dfrac{100-9-10}{81}=1 \in \mathbb{Z}\end{align} Thus it is true for $n=1$. 2. Let it be t... rac{10^{k+1}-9 k-10}{81} \in \mathbb{Z}\cdots (i)\end{align} 3. For $n=k+1$, then we have \begin{align}... \dfrac{10^{k+1}-1}{9}+\dfrac{10^{k+1}-9 k-10}{81}\end{align} Now $\dfrac{10^{k+1}-1}{9} \in \mathbb{Z}
- Question 2 Exercise 7.2
- 4=35 \times 16 a^3 \\ & \Rightarrow T_4=560 a ^3 \end{align} =====Question 2(ii)===== Find the indicat... =-{}^{10}C_7 \,2^{-3}\cdot 3^7 x^3 \cdot y^{-7} \end{align} =====Question 2(iii)===== Find the indi... {21 !}{(21-r) ! r !}(-1)^r \cdot x^{21 \cdot 3 r}\end{align} But the term $T_{r+1}$ independent of $x$ ... ^7 \cdot x^{21-3.7} \\ \Rightarrow T_8&=-^{21}C_7\end{align} Hence $T_8$ is independent of $x$ which
- Question 5 Exercise 7.2
- c{a^4}{x^4} \cdot b^4 x^4 \\ & =70 \cdot a^4 b^4 \end{align} Thus $T_5$ is the middle term of the expan... 30618}{16} x^{13}\\ & T_5=\dfrac{15309}{8} x^{13}\end{align} Putting $r=5$ to get the $2^{\text {nd }}$... }{2^5} \cdot(x^2)^5\\ &=-\dfrac{5103}{16} x^{14} \end{align} Thus the two middle terms are $$T_5=\dfrac... ^5 \cdot y^5 \\ \rightarrow T_6&=-252 x^{10} y^5 \end{align} is the required middle term. ====Go T
- Question 8 Exercise 7.2
- i \quad 7 \\ & =5.82 \quad \because \quad i \\ & \end{aligned} Hence by perpe:ts of like binomial theor... ^{10}$ Now $T_{5} !=\left(\begin{array}{c}10 \\ 5\end{array}\right) 3^{10} 5-2 \gamma^{15}$ we compute ... {aligned} & T_{51}=\left(\begin{array}{c} 0 \\ 5 \end{array}\right) 3^{10 i-5} \quad 2, \frac{3}{4} y \... t)^2 \\ & \frac{25213^{16}}{2^5}-46.5016475 \\ & \end{aligned} $$ Hence $T$ is mumerically greatest le