The number of elements in the set n ∈ 1, 2, 3, , 100 ∣ ( 11 ) ^ n >...

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

$⇒ {11}^{n} - {9}^{n} > {10}^{n}$

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

Using ${\left(a + b\right)}^{n} - {\left(a - b\right)}^{n} = 2\left[C1 n \cdot {a}^{n - 1} \cdot b + C3 n \cdot {a}^{n - 3} \cdot {b}^{3} + C5 n \cdot {a}^{n - 5} \cdot {b}^{5} + . . .\right]$

$⇒ 2\left[C1 n \cdot {10}^{n - 1} + C3 n \cdot {10}^{n - 3} + C5 n \cdot {10}^{n - 5} + ⋯ ⋯\right] > {10}^{n}$

$⇒ 2n \cdot {10}^{n - 1} + 2\left[C3 n \cdot {10}^{n - 3} + C5 n \cdot {10}^{n - 5} + ⋯ ⋯\right] > {10}^{n}$ $\ldots \left(1\right)$

For $n = 5,$ the $L . H . S .$ is $2 \cdot 5 \cdot {10}^{4} + 2\left[C3 5 \cdot {10}^{2} + C5 5\right]$

$⇒ {10}^{5} + 2\left[C3 5 \cdot {10}^{2} + C5 5\right] > {10}^{5}$ which is true.

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

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

$⇒ 2n \cdot {10}^{n - 1} + 2\left[C3 n \cdot {10}^{n - 3} + C5 n \cdot {10}^{n - 5} + ⋯ ⋯\right] > {10}^{n}$

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

For $n = 4,$ the inequality \left(1\right)$is not satisfied.</p><p>Hence, the inequality does not hold good for$n = 1, 2, 3, 4$</p><p>So, required number of elements$= 96 .