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 summation
We start by rewriting the expression inside the limit:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{2^r + 3^r}{6^r} = \lim_{n \to \infty} \sum_{r=1}^{n} \left( \frac{2^r}{6^r} + \frac{3^r}{6^r} \right) \]
This can be simplified to:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \left( \left(\frac{2}{6}\right)^r + \left(\frac{3}{6}\right)^r \right) = \lim_{n \to \infty} \sum_{r=1}^{n} \left( \left(\frac{1}{3}\right)^r + \left(\frac{1}{2}\right)^r \right) \]

### Step 2: Split the summation
Now we can split the summation into two separate sums:
\[ \lim_{n \to \infty} \left( \sum_{r=1}^{n} \left(\frac{1}{3}\right)^r + \sum_{r=1}^{n} \left(\frac{1}{2}\right)^r \right) \]

### Step 3: Identify the geometric series
Both of these sums are geometric series. The formula for the sum of the first \( n \) terms of a geometric series is:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio.

For the first series \( \sum_{r=1}^{n} \left(\frac{1}{3}\right)^r \):
- \( a = \frac{1}{3} \)
- \( r = \frac{1}{3} \)

Thus,
\[ \sum_{r=1}^{n} \left(\frac{1}{3}\right)^r = \frac{\frac{1}{3} \left(1 - \left(\frac{1}{3}\right)^n\right)}{1 - \frac{1}{3}} = \frac{\frac{1}{3} \left(1 - \left(\frac{1}{3}\right)^n\right)}{\frac{2}{3}} = \frac{1}{2} \left(1 - \left(\frac{1}{3}\right)^n\right) \]

For the second series \( \sum_{r=1}^{n} \left(\frac{1}{2}\right)^r \):
- \( a = \frac{1}{2} \)
- \( r = \frac{1}{2} \)

Thus,
\[ \sum_{r=1}^{n} \left(\frac{1}{2}\right)^r = \frac{\frac{1}{2} \left(1 - \left(\frac{1}{2}\right)^n\right)}{1 - \frac{1}{2}} = \frac{\frac{1}{2} \left(1 - \left(\frac{1}{2}\right)^n\right)}{\frac{1}{2}} = 1 - \left(\frac{1}{2}\right)^n \]

### Step 4: Combine the results
Now we combine the results of both sums:
\[ \lim_{n \to \infty} \left( \frac{1}{2} \left(1 - \left(\frac{1}{3}\right)^n\right) + \left(1 - \left(\frac{1}{2}\right)^n\right) \right) \]

### Step 5: Evaluate the limit
As \( n \to \infty \), both \( \left(\frac{1}{3}\right)^n \) and \( \left(\frac{1}{2}\right)^n \) approach 0. Therefore, we have:
\[ \lim_{n \to \infty} \left( \frac{1}{2} \cdot 1 + 1 \right) = \frac{1}{2} + 1 = \frac{3}{2} \]

### Final Answer
Thus, the value of the limit is:
\[ \frac{3}{2} \]