JEE Main 2021 — Mathematics Algebra

We have to check ${11}^{n}>{10}^{n}+{9}^{n}$11n>10n+9n

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

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

Using ${(a+b)}^{n}-{(a-b)}^{n}=2[C1n\cdot {a}^{n-1}\cdot b+C3n\cdot {a}^{n-3}\cdot {b}^{3}+C5n\cdot {a}^{n-5}\cdot {b}^{5}+...]$(a+b)n−(a−b)n=2[C1n⋅an−1⋅b+C3n⋅an−3⋅b3+C5n⋅an−5⋅b5+...]

$$\Rightarrow 2[C1n\cdot {10}^{n-1}+C3n\cdot {10}^{n-3}+C5n\cdot {10}^{n-5}+\cdots \cdots ]>{10}^{n}$$⇒2[C1n⋅10n−1+C3n⋅10n−3+C5n⋅10n−5+⋯⋯]>10n

$\Rightarrow 2n\cdot {10}^{n-1}+2[C3n\cdot {10}^{n-3}+C5n\cdot {10}^{n-5}+\cdots \cdots ]>{10}^{n}$⇒2n⋅10n−1+2[C3n⋅10n−3+C5n⋅10n−5+⋯⋯]>10n $\ldots (1)$…(1)

For $n=5,$n=5, the $L.H.S.$L.H.S. is $2\cdot 5\cdot {10}^{4}+2[C35\cdot {10}^{2}+C55]$2⋅5⋅104+2[C35⋅102+C55]

$\Rightarrow {10}^{5}+2[C35\cdot {10}^{2}+C55]>{10}^{5}$⇒105+2[C35⋅102+C55]>105 which is true.

Now, for $n=6,7,8,\ldots \ldots ,100$n=6,7,8,……,100

$$2n\cdot {10}^{n-1}>{10}^{n}$$2n⋅10n−1>10n

$$\Rightarrow 2n\cdot {10}^{n-1}+2[C3n\cdot {10}^{n-3}+C5n\cdot {10}^{n-5}+\cdots \cdots ]>{10}^{n}$$⇒2n⋅10n−1+2[C3n⋅10n−3+C5n⋅10n−5+⋯⋯]>10n

$\Rightarrow {11}^{n}-{9}^{n}>{10}^{n}$⇒11n−9n>10n for all $n=5,6,7,\ldots \ldots ,100$n=5,6,7,……,100

For $n=4,$n=4, the inequality $(1)$(1) is not satisfied.

Hence, the inequality does not hold good for $n=1,2,3,4$n=1,2,3,4

So, required number of elements$=96.$=96.