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- Question 2 Exercise 7.2
- \frac{7 !}{(7-r) ! r !}(2)^{7-r } a^r $$ For $4^{\text {th }}$ term, putting $r=3$ \begin{align} & T_{3+... ==== Find the indicate term in the expansion $8^{\text {th }}$ term in $(\dfrac{x}{2}-\dfrac{3}{y})^{10}... (\dfrac{x}{2})^{10-r}(-\dfrac{3}{y})^r$$ For $8^{\text {th }}$ term, putting $r=7$ \begin{align}T_{7+1}&... ==== Find the indicate term in the expansion $8^{\text {th }}$ term in $3^{\text {rd }}$ term in $(x^2+\
- Question 11 Exercise 7.3
- \cdot 16=\frac{3}{2} \\ & \Rightarrow 6 n=2 n-2 \text { or } n=-\frac{1}{2} . \\ & \text { Putting } n=-\frac{1}{2} \text { in Eq.(1), we get } \\ & -\frac{1}{2} x=\frac{1}{4} \Rightarrow x=-\frac{1}{2} \text {. Thus } \\ & y+1=\left(1-\frac{1}{2}\right)^{\f
- Question 5 Exercise 7.2
- middle terms that are $(\dfrac{9+1}{2})^{t h}=5^{\text {th }}$ and $(\dfrac{9+3}{2})^{t h}=6^{t h}$ The... } x^{13}\end{align} Putting $r=5$ to get the $2^{\text {nd }}$ middle term that is: \begin{align}T_6&=\d... o middle terms are $$T_5=\dfrac{15309}{8} x^{13} \text { and } T_6=-\dfrac{5103}{16} x^{14}$$ =====Que... is the required middle term. ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:so
- Question 13 Exercise 7.1
- gin{align} & 2^{k+1}=2^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 propo... it is true for all $n \geq 4$. ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-1-p12 |< Question 12 ]]</btn></text> <text align="right"><btn type="success">[[math-1
- Question 7 Exercise 7.2
- dot 5 \cdot 2 \cdot 9 \\ & =64+480+180=692 \\ & \text { Thus }(2+\sqrt{3})^5+(2-\sqrt{3})^5=724\end{ali... 2})^2] \\ & =8 \sqrt{2}[1+2] \\ & =24 \sqrt{2} . \text { Thus } \\ (1+\sqrt{2})^4-(1-\sqrt{2})^4&=24 \sq... a^2 b^3+2 a b^4 . \end{aligned} $$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-2-p6 |< Question 6 ]]</btn></text> <text align="right"><btn type="success">[[math-1
- Question 2 Exercise 7.3
- 000008+\ldots] \\ & -5[1.019808 . .] \cong 5.099 \text {. } \\ & \end{aligned} $$ (ii) $\frac{1}{\sqrt{0... )^{-\frac{1}{2}} $$ $$ =(1-0.002)^{-\frac{1}{2}} \text {. } $$ Using binomial expansion now $$ \begin{a... prox 5.013298 . \end{aligned} ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-3-p1 |< Question 1 ]]</btn></text> <text align="right"><btn type="success">[[math-1
- Question 5 and 6 Exercise 7.3
- in{aligned} & =\left(1-\frac{3 x}{2}+\frac{x}{4}+\text { higher powers of } \mathrm{x}\right) \\ & \time... & =1-\frac{5 x-10 x}{8} \\ & =1-\frac{5 x}{8} . \text { Hence } \end{aligned} $$ $$ \frac{(8+3 x)^{\fra... frac{3}{x}-\frac{9}{2 x^2}$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-3-p4 |< Question 4 ]]</btn></text> <text align="right"><btn type="success">[[math-1
- Question 10 Exercise 7.3
- x=-\frac{1}{4} \\ & \Rightarrow x=\frac{1}{2} . \text { Thus } \\ & \left(1+\frac{1}{2}\right)^{-\frac{... ghtarrow 32 n-15 n=-15 \\ & \Rightarrow 17 n=-15 \text { or } n=-\frac{15}{17} . \end{aligned} $$ Putti... }{7}\right)^{\frac{15}{17}} $$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-3-p7 |< Question 9 ]]</btn></text> <text align="right"><btn type="success">[[math-1
- Question 12 Exercise 7.3
- \cdot 16=\frac{3}{2} \\ & \Rightarrow 6 n=2 n-2 \text { or } n=-\frac{1}{2} . \end{aligned} $$ Putting... \cdot 16=\frac{3}{2} \\ & \Rightarrow 6 n=2 n-2 \text { or } n=-\frac{1}{2} . \end{aligned} $$ Putting... ich is the required result. ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-3-p9 |< Question 11 ]]</btn></text> <text align="right"><btn type="success">[[math-1
- Question 13 Exercise 7.3
- and higher of $x$ can be ignored. Show that $n^{\text {th }}$ root of $1+x$ is equal to $\frac{2 n+(n+1... d} $$ $$ +\ldots=\frac{2 n+(n+1) x}{2 n+(n-1) x} \text {. } $$ Which is the required result. ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-3-p10 |< Question 12 ]]</btn></text> <text align="right"><btn type="success">[[math-1
- Question 6 Exercise 7.1
- +2)-1 \\ & =(k+2)(k+1) !-1 \\ & =(k+2) !-1 \quad \text { as }(k+2)(k+1) !=(k+2) ! \\ \rightarrow 1(1 !)+... for all $n \in \mathbf{N}$. ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-1-p5 |< Question 5 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-1-p7|Question 7 >]]</btn></text>
- Question 7 Exercise 7.1
- t is $a_{k+1}=(k+1)(k+ 2)$. Adding this $(k+1)^{\text {th }}$ term to both sides of the induction hypot... e for all $n \in \mathbf{N}$. ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-1-p6 |< Question 6 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-1-p8|Question 8 >]]</btn></text>
- Question 14 Exercise 7.1
- then $$3^{2 n-1}+2^{2 n-1}=3^{2.1-1}+2^{2.1-1}=5 \text {. }$$ $5$ divides $5$, hence $5$ is a factor of ... for all $n \in \mathbf{N}$. ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-1-p13 |< Question 13 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-1-p15|Question 15 >]]</btn></text
- Question 4 Exercise 7.2
- {align}a^{9-r}b^r&=a^6 b^3\\ \Rightarrow 9 -r=6 \text{or} r=3\end{align} Putting $r=3$ in the above, we... s $-^9C_3 \dfrac{2^6}{27} $ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-2-p3 |< Question 3 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-2-p5|Question 5 >]]</btn></text>
- Question 8 Exercise 7.2
- & \frac{(1+n) i}{1+i} \quad \frac{1+10}{1}{ }^4 \text { here } \\ & x^2=\frac{3}{2} \frac{3}{2}: 8 \\ &... ac{3}{4}$ and is 19375453125 ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-2-p7 |< Question 7 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-2-p9|Question 9 >]]</btn></text>