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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 10 Exercise 7.3
- with the expansion of $(1+x)^n$ that is $$ \begin{aligned} & 1+n x+\frac{n(n-1)}{2 !} x^2 \\ & +\frac{n(n-1(n-2))}{3 !} x^3+\ldots \end{aligned} $$ Comparing both the series, we have $n x=-\... {16}$ Dividing Eq.(2) by Eq.(3), we get $$ \begin{aligned} & \frac{n-1}{2 n}=\frac{3}{32} \cdot 16=\frac{... htarrow 3 n=n-1 \Rightarrow n=-\frac{1}{2} . \end{aligned} $$ Putting $n=-\frac{1}{2}$ in Eq.(1), we get
- 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 $$ Putting $r=6$ in the above $T_{r+1}$ \begin{align}T_{6-1}&=\dfrac{9 !}{(9-6) ! 6 !}\cdot \dfrac{4^{... (2^2)^3 \\ \Rightarrow T_7&=84 \times 27=2268\end{align} Hence $T_7$ is independent of $x$ and is $2268.$
- Question 7 Exercise 7.2
- 5$ ====Solution==== Using binomial formula \begin{align}(2+\sqrt{3})^5+(2 \cdot \sqrt{3})^5& =[(2)^5+{ }^... cdot(\sqrt{3})^4+2{ }^5 C_5 \cdot(\sqrt{3})^5\end{align} simplifing, we get \begin{align} & =2 \cdot 2^5+2^5 C_2 \cdot 2^3 \cdot(\sqrt{3})^2+2^5 C_4 \cdot 2 \cd... ext { Thus }(2+\sqrt{3})^5+(2-\sqrt{3})^5=724\end{align} =====Question 7(ii)===== $(1+\sqrt{2})^4-(1-\sq
- Question 5 and 6 Exercise 7.3
- j^{\frac{2}{3}}}{2 \cdot 3 x+4-5 x} $$ $$ \begin{aligned} & =\frac{8^{\frac{2}{3}}\left(1+\frac{3 x}{8}\... 1-\frac{5 x}{4}\right)^{-\frac{1}{2}}\right] \end{aligned} $$ Applying binomial expansion and neglecting... glecting $x^2$ and higher powers of $x$ $$ \begin{aligned} & =\left(1-\frac{3 x}{2}+\frac{x}{4}+\text { h... c{5 x}{4}\right)\left(1+\frac{5 x}{8}\right) \end{aligned} $$ Multiplying and neglecting $x^2$ and highe
- Question 12 Exercise 7.3
- with the expansion of $(1+x)^n$ that is $$ \begin{aligned} & 10+n x+\frac{n(n-1)}{2 !} x^2+ \\ & \frac{n(n-1(n-2))}{3 !} x^3+\ldots \end{aligned} $$ Comparing both the series, we have $n x=\f... {16}$ Dividing Eq.(2) by Eq.(3), we get $$ \begin{aligned} & \frac{n-1}{2 n}=\frac{1.3}{2 !} \cdot \frac{... rrow 6 n=2 n-2 \text { or } n=-\frac{1}{2} . \end{aligned} $$ Putting $n=-\frac{1}{2}$ in Eq.(1), we get
- Question 1 Exercise 7.3
- binomial theorem to tind the four terms $$ \begin{aligned} & (1-x)^{\frac{1}{2}}=1+\frac{1}{2} x+ \\ & \f... 1}{2}\left(-\frac{1}{2}-1\right)}{2 !}(-x)^2 \end{aligned} $$ $$ \begin{aligned} & +\frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3 ... \frac{3}{8} x^2+\frac{1}{16} x^3+\ldots \\ & \end{aligned} $$ Solution: Using binomial theorem $$ \begin
- Question 10 Exercise 7.2
- gin{array}{l}n \\ 2\end{array}\right)$. $$ \begin{aligned} & \left(\begin{array}{l} n \\ 3 \end{array}\ri... \begin{array}{l} n \\ n \end{array}\right) . \end{aligned} $$ which shows that the sum of the :nefficiens is $?^n$. Now we know that $$ \begin{aligned} & (1+x)^n=\left(\begin{array}{l} n \\ 0 \end{a... array}{c} n \\ n \end{array}\right) 1^n \\ & \end{aligned} $$ If we put $x=-1$ in the above eyuation, we
- Question 2 Exercise 7.3
- (i) $\sqrt{26}$ Solution: We are given $$ \begin{aligned} & \sqrt{26}=\sqrt{25+1} \\ & =\sqrt{25} \sqrt{... }=5\left[1+\frac{1}{25}\right]^{\frac{1}{2}} \end{aligned} $$ Using binomial expansion $$ \begin{aligned} & \sqrt{26}=5\left[1+\frac{1}{25}\right]^{\frac{1}{2}... -5[1.019808 . .] \cong 5.099 \text {. } \\ & \end{aligned} $$ (ii) $\frac{1}{\sqrt{0.998}}$ Solution: We
- Question 12 Exercise 7.1
- \mathbb{Z}$$ 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 9 Exercise 7.3
- }{1-x}\right)^2$. Solution: Given that: $$ \begin{aligned} & \left(\frac{1+x}{1-x}\right)^2=(1+x)^2(1-x)^{-2} \\ & =\left(x^2+2 x+1\right)(1-x)^2 \end{aligned} $$ Applying binomial theorem $$ \begin{aligned} & =\left(x^2+2 x+1\right)[1+2 x+ \\ & \frac{-2(-2-1)... t[1+2 x+3 x^2+4 x^3\right. \\ & +\ldots . .] \end{aligned} $$ Generalizing up-to $x^{\prime t}$ as $$ \be
- Question 13 Exercise 7.3
- n+(n-1) x}$ Solution: We have show that $$ \begin{aligned} & (1+x)^{\frac{1}{n}}=\frac{2 n+(n+1) x}{2 n+(... }\left(\frac{1}{n}-1\right)}{2 !} x^2+\ldots \end{aligned} $$ We are considering $$ \begin{aligned} & \frac{2 n+(n+1) x}{2 n+(n-1) x}=\frac{2 n\left[1+\left(\fr... 1+\left(\frac{n-1}{2 n}\right) x\right]^{-1} \end{aligned} $$ Applying binomial theorem now $$ \begin{al
- Question 7 & 8 Review Exercise 7
- pothesis. (3.) For $n=k+1$ then we have $$ \begin{aligned} & 7^{k+1}-3^{k+1}=7.7^k-3.3^k \\ & =(4+3) \cdot 7^k-3.3^k \\ & =4.7^k+3.7^k-3.3^k \end{aligned} $$ $$ \begin{aligned} & =4.7^k+3\left[7^k-3^k\right] \\ & \Rightarrow 7^{k+1}-3^{k+1}=4.7^k+3.4 Q \end{aligned} $$ by induction hypothesis $$ \begin{aligned}