The number of elements in the set

Question:

The number of elements in the set $\{n \in\{1,2,3, \ldots . .100\} \mid$ $\left.(11)^{\mathrm{n}}>(10)^{\mathrm{n}}+(9)^{\mathrm{n}}\right\}$ is__________.

Solution:

$11^{n}>10^{n}+9^{n}$

$\Rightarrow 11^{\mathrm{n}}-9^{\mathrm{n}}>10^{\mathrm{n}}$

$\Rightarrow(10+1)^{\mathrm{n}}-(10-1)^{\mathrm{n}}>10^{\mathrm{n}}$

$\Rightarrow\left\{{ }^{n} C_{1} \cdot 10^{n-1}+{ }^{n} C_{3} 10^{n-0}+{ }^{n} C_{5} 10^{n-5}+\cdots \cdots\right\}>10^{n}$

$\Rightarrow 2 \mathrm{n} \cdot 10^{\mathrm{n}-1}+2\left\{{ }^{\mathrm{n}} \mathrm{C}_{3} 10^{\mathrm{n}-3}+{ }^{\mathrm{n}} \mathrm{C}_{5} 10^{\mathrm{n}-5}+\cdots \cdots\right\}>10^{\mathrm{n}}$ ..........(1)

For $n=5$

$10^{5}+2\left\{{ }^{5} \mathrm{C}_{3} 10^{2}+{ }^{5} \mathrm{C}_{5}\right\}>10^{5}$ (True)

For $\mathrm{n}=6,7,8, \ldots \ldots 100$

$2 \mathrm{n} 10^{\mathrm{n}-1}>10^{\mathrm{n}}$

$\Rightarrow 2 \mathrm{n} 10^{\mathrm{n}-1}+2\left\{{ }^{\mathrm{n}} \mathrm{C}_{3} 10^{\mathrm{n}-3}+{ }^{\mathrm{n}} \mathrm{C}_{5} 10^{\mathrm{n}-5}+\cdots \cdots\right\}>10^{\mathrm{n}}$

$\Rightarrow 11^{\mathrm{n}}-9^{\mathrm{n}}>10^{\mathrm{n}}$ For $\mathrm{n}=5,6,7, \ldots . .100$

For $n=4$, Inequality (1) is not satisfied

$\Rightarrow$ Inequality does not hold good for

$\mathrm{N}=1,2,3,4$

So, required number of elements $=96$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now