Binomial Theorem

Q1. Expand the expression $(1-2x)^5$.

Sol. By the binomial theorem,

Here, $(1-2x)^5=[1+(-2x)]^5$

So $a=1$, $b=-2x$ and $n=5$.

Substituting the values,

Q2. Expand the expression $\left(\frac{2}{x}-\frac{x}{2}\right)^5$.

Sol. By the binomial theorem,

Here, $\left(\frac{2}{x}-\frac{x}{2}\right)^5=\left[\frac{2}{x}+\left(-\frac{x}{2}\right)\right]^5$

So $a=\frac{2}{x}$, $b=-\frac{x}{2}$ and $n=5$.

Substituting the values,

Q3. Expand the expression $(2x-3)^6$.

Sol. By the binomial theorem,

Here, $(2x-3)^6=[2x+(-3)]^6$,

So $a=2x$, $b=-3$ and $n=6$.

Substituting the values,

Q4. Expand the expression $\left(\frac{x}{3}+\frac{1}{x}\right)^5$.

Sol. By the binomial theorem,

Here, $a=\frac{x}{3}$, $b=\frac{1}{x}$ and $n=5$.

Substituting the values,

Q5. Expand the expression $\left(x+\frac{1}{x}\right)^6$.

Sol. By the binomial theorem,

Here, $a=x$, $b=\frac{1}{x}$ and $n=6$.

Substituting the values,

Q6. Using binomial theorem, evaluate $(96)^3$.

Sol. We write $96=100+(-4)$, so $(96)^3=[100+(-4)]^3$.

By the binomial theorem,

Here, $a=100$, $b=-4$ and $n=3$.

Substituting the values,

Q7. Using binomial theorem, evaluate $(102)^5$.

Sol. We write $102=100+2$, so $(102)^5=(100+2)^5$.

By the binomial theorem,

Here, $a=100$, $b=2$ and $n=5$.

Substituting the values,

Q8. Using binomial theorem, evaluate $(101)^4$.

Sol. We write $101=100+1$, so $(101)^4=(100+1)^4$.

By the binomial theorem,

Here, $a=100$, $b=1$ and $n=4$.

Substituting the values,

Q9. Using binomial theorem, evaluate $(99)^5$.

Sol. We write $99=100+(-1)$, so $(99)^5=[100+(-1)]^5$.

By the binomial theorem,

Here, $a=100$, $b=-1$ and $n=5$.

Substituting the values,

Q10. Using Binomial Theorem, indicate which number is larger $(1.1)^{10000}$ or 1000.

Sol. We write $1.1=1+0.1$, so $(1.1)^{10000}=(1+0.1)^{10000}$.

By the binomial theorem,

Here, $a=1$, $b=0.1$ and $n=10000$.

Substituting the values,

Since all the remaining terms are positive,

Hence, $(1.1)^{10000}$ is the larger number.

Q11. Find $(a+b)^4-(a-b)^4$. Hence, evaluate $(\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4$.

Sol. By the binomial theorem,

Here, $n=4$.

Subtracting,

For $(\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4$,

Here, $a=\sqrt{3}$ and $b=\sqrt{2}$.

Substituting the values,

Q12. Find $(x+1)^6+(x-1)^6$. Hence or otherwise evaluate $(\sqrt{2}+1)^6+(\sqrt{2}-1)^6$.

Sol. By the binomial theorem,

Expanding $(x+1)^6$,

Similarly,

Adding,

For $(\sqrt{2}+1)^6+(\sqrt{2}-1)^6$,

Putting $x=\sqrt{2}$,

Q13. Show that $9^{n+1}-8n-9$ is divisible by 64, whenever $n$ is a positive integer.

Sol. We write $9=1+8$, so

By the binomial theorem,

Here, $a=1$, $b=8$ and $n$ is replaced by $n+1$.

Substituting the values,

where $k$ is an integer.

Hence, $9^{n+1}-8n-9$ is divisible by 64.

Q14. Prove that $\sum_{r=0}^{n}3^r\,{}^nC_r=4^n$.

Sol. By the binomial theorem,

To prove,

Comparing (1) and (2), $b=3$ and $a+b=4$.

Putting $a=1$ and $b=3$ in (1),