The value of `lim_(nrarroo)Sigma_(r=1)^(n)(2^(r)+3^(r))/(6^(r))` is equal to

Options

A. 0
B. 1
C. 6
D. `(3)/(2)`

Answer

To solve the limit

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{2^r + 3^r}{6^r}, \]

we can break it down step by step.

### Step 1: Rewrite the limit
We can express the limit as:

\[ s = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{2^r + 3^r}{6^r}. \]

### Step 2: Separate the terms in the summation
We can separate the summation into two parts:

\[ s = \lim_{n \to \infty} \left( \sum_{r=1}^{n} \frac{2^r}{6^r} + \sum_{r=1}^{n} \frac{3^r}{6^r} \right). \]

### Step 3: Simplify the fractions
Now, we can simplify each term:

\[ \frac{2^r}{6^r} = \left(\frac{2}{6}\right)^r = \left(\frac{1}{3}\right)^r, \]
\[ \frac{3^r}{6^r} = \left(\frac{3}{6}\right)^r = \left(\frac{1}{2}\right)^r. \]

### Step 4: Rewrite the limit with simplified terms
Thus, we can rewrite \(s\) as:

\[ s = \lim_{n \to \infty} \left( \sum_{r=1}^{n} \left(\frac{1}{3}\right)^r + \sum_{r=1}^{n} \left(\frac{1}{2}\right)^r \right). \]

### Step 5: Recognize the geometric series
Both summations are geometric series. The sum of a geometric series can be calculated using the formula:

\[ \sum_{r=1}^{\infty} ar^r = \frac{a}{1 - r}, \]

where \(a\) is the first term and \(r\) is the common ratio.

### Step 6: Calculate the first geometric series
For the first series:

\[ \sum_{r=1}^{\infty} \left(\frac{1}{3}\right)^r = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}. \]

### Step 7: Calculate the second geometric series
For the second series:

\[ \sum_{r=1}^{\infty} \left(\frac{1}{2}\right)^r = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1. \]

### Step 8: Combine the results
Now, we can combine the results of both series:

\[ s = \frac{1}{2} + 1 = \frac{1}{2} + \frac{2}{2} = \frac{3}{2}. \]

### Final Result
Thus, the value of the limit is:

\[ \boxed{\frac{3}{2}}. \]