Cho P = 3^0 + 3^1 + 3^2 + 3^3 + ... + 3^99

To solve the given problem, let's start by analyzing the series and the questions provided.

Given:
\[ P = 3^0 + 3^1 + 3^2 + 3^3 + \ldots + 3^{99} \]

This is a geometric series with the first term \( a = 3^0 = 1 \) and the common ratio \( r = 3 \). The sum of the first \( n \) terms of a geometric series is given by:
\[ S_n = a \frac{r^n - 1}{r - 1} \]

For this series:
\[ P = \frac{3^{100} - 1}{3 - 1} = \frac{3^{100} - 1}{2} \]

### Part (a)
Write \( 2P + 1 \) in the form of a power of 3.

\[ 2P + 1 = 2 \left( \frac{3^{100} - 1}{2} \right) + 1 = 3^{100} - 1 + 1 = 3^{100} \]

So, \( 2P + 1 = 3^{100} \).

### Part (b)
Find \( x \in \mathbb{N} \) such that \( 2P + 1 = 3^{x+5} \).

From part (a), we have:
\[ 2P + 1 = 3^{100} \]

We need to find \( x \) such that:
\[ 3^{100} = 3^{x+5} \]

Since the bases are the same, we can equate the exponents:
\[ 100 = x + 5 \]

Solving for \( x \):
\[ x = 100 - 5 \]
\[ x = 95 \]

So, \( x = 95 \).