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- Question 7 & 8 Review Exercise 7
- ion: We using mathematical induction to prove the given statement. (1.) For $n=1$ then $7^k-3^k=7-4=4$. Thus 4 divides 4. Hence given is true for $n=1$. (2.) Let it be true for $n=k>1... is divisible by } 4 . \end{aligned} $$ Thus the given statement is true for $n=k+1$. Hence it is true f... +1} \geq[1+(k+1) x] . \end{aligned} $$ Hence the given is true for $n=k+1$. Thus by mathematical inducti
- Question 11 Exercise 7.3
- s$ then show that $y^2+2 y-1=0$. Solution: We are given $y=\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2
- Question 10 Exercise 7.3
- 3}{2 !} \cdot \frac{1}{2^4}+\ldots$ Solution: The given series is binomial series. Let it be identical wi... }+\frac{5.8 .11}{8.12 .16}+\ldots$ Solution: The given series is binomial series. Let it be identical wi... }+\ldots . \end{aligned} $$ Hence the sum of the given series is: $$ \left(\frac{24}{7}\right)^{\frac{15
- Question 9 Exercise 7.3
- }$ in $\left(\frac{1+x}{1-x}\right)^2$. Solution: Given that: $$ \begin{aligned} & \left(\frac{1+x}{1-x}\
- Question 7 and 8 Exercise 7.3
- d $b$. Solution: We are taking L.H.S of the above given equation and apply the binomial theorem $$ \begin... We are taking numerator in the L.H.S of the above given equation ====Go To==== <text align="left"><b
- Question 4 Exercise 7.3
- frac{1-3 x}{1+4 x}}=1-\frac{7 x}{2} $$ Solution: Given that $$ \sqrt{\frac{1-3 x}{1-4 x}}=(1-3 x)^{\frac
- Question 5 and 6 Exercise 7.3
- +3 x) \sqrt{4-5 x}}=1-\frac{5 x}{8} $$ Solution: Given that: $$ \frac{\sqrt[4]{3}-3 x j^{\frac{2}{3}}}{2... c{x \sqrt{x^2-2 x}}{(x+1)^2} $$ Solution: We are given $$ \begin{aligned} & \frac{x \sqrt{x^2-2 x}}{(x+1
- Question 3 Exercise 7.3
- d $\sqrt{\frac{1-x}{1+x}}$ up-to $x^3$. Solution: Given $\sqrt{\frac{1-x}{1+x}}$ $$ =(1-x)^{\frac{1}{2}}(
- Question 2 Exercise 7.3
- decimal places. (i) $\sqrt{26}$ Solution: We are given $$ \begin{aligned} & \sqrt{26}=\sqrt{25+1} \\ & =... $ (ii) $\frac{1}{\sqrt{0.998}}$ Solution: We are given that $$ \frac{1}{\sqrt{0.998}}=(0.998)^{-\frac{1}
- Question 12 Exercise 7.3
- then show that $4 y^2+4 y-1=0$. Solution: We are given $$ 2 y=\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{
- Question 6 Exercise 7.2
- t is independent of $x$. The general term of the given expansion is: \begin{align} T_{r+1}&=\dfrac{23 !}
- Question 5 Exercise 7.2
- c{8+2}{2})^{t h}=5^{t h}$. Now $T_{r+1}$ of the given expansion is: $$T_{r+1}=\dfrac{8 !}{(8-r) ! r !}(... c{10+2}{2})^{t h}=6^{t h}$. Now $T_{r+1}$ of the given expansion is: $$T_{r+1}=\dfrac{10 !}{(10-r) ! r !
- Question 3 Exercise 7.2
- t $T_{r+1}$ be the term independent of $x$ in the given expansion. $T_{r+1}$ of the given expansion is: \begin{align}T_{r+1}&=\dfrac{9 !}{(9-r) ! r !}(\dfrac{4... t $T_{r+1}$ be the term independent of $x$ in the given expansion. $T_{r+1}$ of the given expansion is: \begin{align} T_{r+1}&=\dfrac{10 !}{(10-r) ! r !}(x)^{1
- Question 2 Exercise 7.2
- $, $a=2$ and $b=a$. Thus the general term of the given expansion is: $$T_{r+1}=\frac{7 !}{(7-r) ! r !}(2... $b=-\dfrac{3}{y}$. Thus the general term of the given expansion is: $$T_{r+1}=\dfrac{10 !}{(10-r) ! r !... t $T_{r+1}$ be the term independent of $x$ in the given expansion. $T_{r+1}$ of the given expansion is: \begin{align}T_{r-1}&=\dfrac{21 !}{(21-r) ! r !}(x)^{21
- Question 5 Exercise 7.1
- d{aligned} 3. Now $n=k+1$ the $(k+1)$ term of the given series on left is $a_{k+1}=(k+1)^3$. Adding this