Question

Let $f(n) = 2n + 1$, $g(n) = 1 + (n + 1)^{2n}$ for all $n \in \mathbb{N}$. Then

• A. f(n)>g(n)
• B. f(n)<g(n)
• C. f(n) and g(n) are not comparable
• D. f(n)>g(n) if n be even and f(n)<g(n) if n be odd.

Answer 1

The Correct Option is B

We are given two functions: \[ f(n) = 2n + 1 \] and \[ g(n) = 1 + (n + 1)^{2n} \]

Step 1: Compare the growth of \( f(n) \) and \( g(n) \)
- The function \( f(n) = 2n + 1 \) is a linear function, which grows linearly as \( n \) increases. - The function \( g(n) = 1 + (n + 1)^{2n} \) is much more complicated, but it's important to note that \( (n + 1)^{2n} \) grows exponentially with \( n \), which is much faster than the linear growth of \( f(n) \). Thus, for large values of \( n \), the exponential term in \( g(n) \) will dominate, making \( g(n) \) much larger than \( f(n) \). 

Step 2: Compare for small values of \( n \)
- For \( n = 1 \): \[ f(1) = 2(1) + 1 = 3, \quad g(1) = 1 + (1 + 1)^{2(1)} = 1 + 2^2 = 5 \] So, \( f(1) < g(1) \). - For \( n = 2 \): \[ f(2) = 2(2) + 1 = 5, \quad g(2) = 1 + (2 + 1)^{2(2)} = 1 + 3^4 = 1 + 81 = 82 \] So, \( f(2) < g(2) \). As we can see, for small values of \( n \), \( f(n) < g(n) \), and for large values of \( n \), \( g(n) \) grows exponentially faster than \( f(n) \). 

Step 3: Conclusion
Since \( g(n) \) grows much faster than \( f(n) \) as \( n \) increases, we conclude that: \[ f(n) < g(n) \quad \text{for all} \, n \in \mathbb{N} \]

Answer:

\[ \boxed{f(n) < g(n)} \]