For all `n in N, n^(4)` is less than

Options

A. `10^(n)`
B. `4^(n)`
C. 4n
D. `10^(10)`

Answer

To prove that for all \( n \in \mathbb{N} \), \( n^4 < 10^n \), we will use mathematical induction.

### Step 1: Base Case
We start with the base case \( n = 1 \).
\[ 1^4 = 1 \quad \text{and} \quad 10^1 = 10 \]
Since \( 1 < 10 \), the base case holds true.

### Step 2: Inductive Hypothesis
Assume that the statement is true for some \( n = k \), i.e., assume that:
\[ k^4 < 10^k \]

### Step 3: Inductive Step
We need to prove that the statement is also true for \( n = k + 1 \), i.e., we need to show:
\[ (k + 1)^4 < 10^{k + 1} \]

We can expand \( (k + 1)^4 \) using the binomial theorem:
\[ (k + 1)^4 = k^4 + 4k^3 + 6k^2 + 4k + 1 \]

Now, we need to compare \( k^4 + 4k^3 + 6k^2 + 4k + 1 \) with \( 10^{k + 1} \):
\[ 10^{k + 1} = 10 \cdot 10^k \]

From our inductive hypothesis, we know:
\[ k^4 < 10^k \]

Thus, we can substitute \( k^4 \) in our inequality:
\[ k^4 + 4k^3 + 6k^2 + 4k + 1 < 10^k + 4k^3 + 6k^2 + 4k + 1 \]

Now, we need to show that:
\[ 10^k + 4k^3 + 6k^2 + 4k + 1 < 10 \cdot 10^k \]

This simplifies to:
\[ 4k^3 + 6k^2 + 4k + 1 < 9 \cdot 10^k \]

### Step 4: Verifying the Inequality
We can check this inequality for \( k = 1 \) and \( k = 2 \) to see if it holds:

For \( k = 1 \):
\[ 4(1)^3 + 6(1)^2 + 4(1) + 1 = 4 + 6 + 4 + 1 = 15 \]
\[ 9 \cdot 10^1 = 90 \]
Since \( 15 < 90 \), the inequality holds.

For \( k = 2 \):
\[ 4(2)^3 + 6(2)^2 + 4(2) + 1 = 32 + 24 + 8 + 1 = 65 \]
\[ 9 \cdot 10^2 = 900 \]
Since \( 65 < 900 \), the inequality holds.

### Conclusion
By the principle of mathematical induction, since the base case holds and the inductive step has been verified, we conclude that:
\[ n^4 < 10^n \quad \text{for all} \quad n \in \mathbb{N} \]