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- Question 1 Exercise 7.2
- =====Question 1(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 ... =====Question 1(ii)===== Expand by using Binomial theorem: $(1+x y)^7$ ====Solution==== Using binonial theorem \begin{align} & (1+x y)^7=1+{ }^7 C_1(1)^6(x y)+{ }^7
- Question 1 Exercise 7.3
- i) (1 - x) $\frac{1}{2}$ Solution: Using binomial theorem to tind the four terms $$ \begin{aligned} & (1-x)... s \\ & \end{aligned} $$ Solution: Using binomial theorem $$ \begin{aligned} & (1-x)^{\frac{3}{2}}=1-\frac{
- Question 7 Exercise 7.2
- (iii) $(a+b)^5+(a-b)^5$ Solution: Using binomial theorem $$ \begin{aligned} (a+b)^5+(a-b)^5&=\left[\left(\
- Question 8 Exercise 7.2
- \end{aligned} Hence by perpe:ts of like binomial theorem, we hisu that: $p+1$ - 5.. 1 - 6 icrm is numerica
- Question 3 Exercise 7.3
- )^{-\frac{1}{2}} \text {. } $$ Applying binomial theorem, $$ \begin{aligned} & (1-x)^{\frac{1}{2}}(1+x)^{\
- Question 5 and 6 Exercise 7.3
- }\right)^{-2} \end{aligned} $$ Applying binomial theorem and neglecting $\frac{1}{x^3}$ etc $$ \begin{alig
- Question 7 and 8 Exercise 7.3
- f the above given equation and apply the binomial theorem $$ \begin{aligned} & (1+x)^{\frac{1}{4}}+(1-x)^{\
- Question 9 Exercise 7.3
- right)(1-x)^2 \end{aligned} $$ Applying binomial theorem $$ \begin{aligned} & =\left(x^2+2 x+1\right)[1+2
- Question 13 Exercise 7.3
- x\right]^{-1} \end{aligned} $$ Applying binomial theorem now $$ \begin{aligned} & {\left[1+\left(\frac{n+1
- Question 14 Exercise 7.3
- x^p-q x^q=p(1+h)^p-q(1+h)^q $$ Applying binomial theorem on the R.H.S of the above last equation, $$ \begi
- Question 5 & 6 Review Exercise 7
- n write $$ (0.99)^5=(1-0.01)^5 $$ Using binomial theorem, we have $$ \begin{aligned} & (1-0.01)^5 \cong{ }