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- Question 1 Exercise 7.3
- irst four terms in the expansion of (i) (1 - x) $\frac{1}{2}$ Solution: Using binomial theorem to tind the four terms $$ \begin{aligned} & (1-x)^{\frac{1}{2}}=1+\frac{1}{2} x+ \\ & \frac{\frac{1}{2}\left(-\frac{1}{2}-1\right)}{2 !}(-x)^2 \end{aligned} $$ $$ \begin{align
- Question 5 and 6 Exercise 7.3
- igher of $x$ may be negleeled. then show that $$ \frac{(8+3 x)^{\frac{2}{3}}}{(2+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 \cdot 3 x+4-5 x} $$ $$ \b
- Question 2 Exercise 7.3
- & \sqrt{26}=\sqrt{25+1} \\ & =\sqrt{25} \sqrt{1+\frac{1}{25}}=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\left[1+\frac{
- Question 10 Exercise 7.3
- Q10 Find the sum of the following series: (i) $1-\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2^4}+\ldots$ Solution: The given series is binomial series. Let it be identica... of $(1+x)^n$ that is $$ \begin{aligned} & 1+n x+\frac{n(n-1)}{2 !} x^2 \\ & +\frac{n(n-1(n-2))}{3 !} x^
- Question 12 Exercise 7.3
- (KPTB or KPTBB) Peshawar, Pakistan. Q12 If $2 y=\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2^4}+\frac{1.3 \cdot 5}{3 !} \cdot \frac{1}{2^6}+\ldots$ then show that $4 y^2+4 y-1=0$. Solution: W
- Question 13 Exercise 7.3
- hat $n^{\text {th }}$ root of $1+x$ is equal to $\frac{2 n+(n+1) x}{2 n+(n-1) x}$ Solution: We have show that $$ \begin{aligned} & (1+x)^{\frac{1}{n}}=\frac{2 n+(n+1) x}{2 n+(n-1) x} \\ & \frac{2 n+(n+1) x}{2 n+(n-1) x} \\ & =1+\frac{1}{n} x+\frac{\frac{1}{n}\
- Question 9 Exercise 7.1
- the formulas below by mathematical induction, $\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ldots+\dfrac{1}{3^n}=\dfrac{1}{2}[1-\dfrac{1}{3^n}]$ ====Solution==== 1. For $n=1$ then $$\dfra
- Question 3 Exercise 7.3
- or KPTBB) Peshawar, Pakistan. Q3 Expand $\sqrt{\frac{1-x}{1+x}}$ up-to $x^3$. Solution: Given $\sqrt{\frac{1-x}{1+x}}$ $$ =(1-x)^{\frac{1}{2}}(1+x)^{-\frac{1}{2}} \text {. } $$ Applying binomial theorem, $$ \begin{aligned} & (1-x)^{\frac{
- Question 1 Exercise 7.2
- i)===== Expand by using Binomial theorem: $(x^2-\dfrac{1}{y})^4$ ====Solution==== Using binomial theorem \begin{align}(x^2-\dfrac{1}{y})^4&=(x^2)^4+{ }^4 C_1(x^2)^3(-\dfrac{1}{y})+ \\ & { }^4 C_2(x^2)^2(-\dfrac{1}{y})^2+{ }^4 C_3(x^2)(-\dfrac{1}{y})^3 + { }^4 C_4(-\dfrac{1}{y
- Question 7 and 8 Exercise 7.3
- x^4$ and higher powers are neglected and $(1-x)^{\frac{1}{4}}+(1-x)^{\frac{1}{4}}=a-b x^2$ then find $a$, and $b$. Solution: We are taking L.H.S of the above ... the binomial theorem $$ \begin{aligned} & (1+x)^{\frac{1}{4}}+(1-x)^{\frac{1}{4}} \\ & =\left[1+\frac{x}{4}+\frac{\frac{1}{4}\left(\frac{1}{4}-1\right)}{2 !}
- Question 3 Exercise 7.2
- the term independent of $x$ in the expansion $(\dfrac{4 x^2}{3}-\dfrac{3}{2 x})$ ====Solution==== In the above expansion $n=9, \quad a=\dfrac{4 x^2}{3}$ and $b=-\dfrac{3}{2 x}$. Let $T_{r+1}$ be the term independent of $x$ in the given expansio
- Question 11 Exercise 7.3
- d (KPTB or KPTBB) Peshawar, Pakistan. Q11 If $y=\frac{1}{2^2}+\frac{1.3}{2 !} \cdot \frac{1}{2^4}+\frac{1 \cdot 3 \cdot 5}{3 !} \cdot \frac{1}{2^6}+\ldots$ then show that $y^2+2 y-1=0$. Solutio
- Question 5 Exercise 7.2
- (i)===== Find middle term in the expansion of $(\dfrac{a}{x}+b x)^8$ ====Solution==== Since we see that $a=\dfrac{a}{x}$. $b=b x$ and $n=8$ Since $n-8$ is a the ... are $8+1=9$ The middle term is only one so, $$(\dfrac{8+2}{2})^{t h}=5^{t h}$. Now $T_{r+1}$ of the given expansion is: $$T_{r+1}=\dfrac{8 !}{(8-r) ! r !}(\dfrac{a}{x})^{8-r}(b x)^r$$ To
- Question 12 Exercise 7.1
- 12(i)===== Show by mathematical induction that $\dfrac{5^{2 n}-1}{24}$ is an integer. Solution: 1. For $n=1$, then $$\dfrac{5^{2 n}-1}{24}=\dfrac{5^{2.1}-1}{24}=\dfrac{24}{24}=1 \in \mathbb{Z}$$ Thus it is true for $n=1$ 2. Let it be true for $n=k>
- Question 4 Exercise 7.2
- taining $x^{23}$ that is: \begin{align}T_{r-1}&=\dfrac{20 !}{(20-r) ! r !}(x^2)^{20 r}(-x)^r \\ & =\dfrac{20 !}{(20-r) ! r !}(-1)^r \cdot x^{40-2 r+r} \\ & =\dfrac{20 !}{(20-r) ! r !}(-1)^r x^{40-r}\end{align} But... 17$. in $T_{r+1}$ we get \begin{align}T_{17-1}&=\dfrac{20 !}{(20 \cdot 17) ! 17 !}(-1)^{17} x^{40-17} \\