`p=sum_(n=0)^(oo) (x^(3n))/((3n)!) , q=sum_(n=1)^(oo) (x^(3n-2))/((3n-2)!), r = sum_(n=1)^(oo) (x^(3n-1))/((3n-1)!)` then p + q + r =

Options

A. `e^x`
B. `e^(2x)`
C. `e^(3x)`
D. `e^(2)`

Answer

To solve the problem, we need to evaluate the sums \( p \), \( q \), and \( r \) and then find \( p + q + r \).

### Step 1: Evaluate \( p \)

The series \( p \) is defined as:

\[ p = \sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!} \]

This series includes terms where \( n \) takes values starting from 0. The first few terms are:

- For \( n = 0 \): \( \frac{x^0}{0!} = 1 \)
- For \( n = 1 \): \( \frac{x^3}{3!} \)
- For \( n = 2 \): \( \frac{x^6}{6!} \)
- For \( n = 3 \): \( \frac{x^9}{9!} \)

Thus, we can express \( p \) as:

\[ p = 1 + \frac{x^3}{3!} + \frac{x^6}{6!} + \frac{x^9}{9!} + \ldots \]

### Step 2: Evaluate \( q \)

The series \( q \) is defined as:

\[ q = \sum_{n=1}^{\infty} \frac{x^{3n-2}}{(3n-2)!} \]

This series starts from \( n = 1 \). The first few terms are:

- For \( n = 1 \): \( \frac{x^{1}}{1!} \)
- For \( n = 2 \): \( \frac{x^{4}}{4!} \)
- For \( n = 3 \): \( \frac{x^{7}}{7!} \)

Thus, we can express \( q \) as:

\[ q = \frac{x^1}{1!} + \frac{x^4}{4!} + \frac{x^7}{7!} + \ldots \]

### Step 3: Evaluate \( r \)

The series \( r \) is defined as:

\[ r = \sum_{n=1}^{\infty} \frac{x^{3n-1}}{(3n-1)!} \]

This series also starts from \( n = 1 \). The first few terms are:

- For \( n = 1 \): \( \frac{x^{2}}{2!} \)
- For \( n = 2 \): \( \frac{x^{5}}{5!} \)
- For \( n = 3 \): \( \frac{x^{8}}{8!} \)

Thus, we can express \( r \) as:

\[ r = \frac{x^2}{2!} + \frac{x^5}{5!} + \frac{x^8}{8!} + \ldots \]

### Step 4: Combine \( p \), \( q \), and \( r \)

Now, we can combine \( p \), \( q \), and \( r \):

\[ p + q + r = \left( 1 + \frac{x^3}{3!} + \frac{x^6}{6!} + \ldots \right) + \left( \frac{x^1}{1!} + \frac{x^4}{4!} + \frac{x^7}{7!} + \ldots \right) + \left( \frac{x^2}{2!} + \frac{x^5}{5!} + \frac{x^8}{8!} + \ldots \right) \]

When we combine these series, we notice that we are summing all terms of the form \( \frac{x^n}{n!} \) for \( n \geq 0 \). This is the Taylor series expansion for \( e^x \):

\[ p + q + r = e^x \]

### Final Result

Thus, we conclude that:

\[ p + q + r = e^x \]