The value of `lim_(n rarr infty )n{(1)/(3n^(2)+8n+4)+(1)/(3n^(2) +16n+16)+...+n "terms"}` is equal to

Options

A. `(1)/(2) log((9)/(5))`
B. `(1)/(3) log ((9)/(5))`
C. `(1)/(4) log ((9)/(5))`
D. None of these

Answer

To find the limit

\[ \lim_{n \to \infty} n \left( \frac{1}{3n^2 + 8n + 4} + \frac{1}{3n^2 + 16n + 16} + \ldots + \frac{1}{3n^2 + 8n + 4n} \right) \]

we can express this sum in a more manageable form.

### Step 1: Rewrite the sum
The sum consists of \( n \) terms, where the \( r \)-th term can be expressed as:

\[ \frac{1}{3n^2 + 8rn + 4r^2} \]

Thus, we can rewrite the limit as:

\[ \lim_{n \to \infty} n \sum_{r=1}^{n} \frac{1}{3n^2 + 8rn + 4r^2} \]

### Step 2: Factor out \( n^2 \)
Next, we factor \( n^2 \) from the denominator:

\[ \frac{1}{3n^2 + 8rn + 4r^2} = \frac{1/n^2}{3 + \frac{8r}{n} + \frac{4r^2}{n^2}} \]

This gives us:

\[ \lim_{n \to \infty} n \sum_{r=1}^{n} \frac{1/n^2}{3 + \frac{8r}{n} + \frac{4r^2}{n^2}} = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} \cdot \frac{1}{3 + \frac{8r}{n} + \frac{4r^2}{n^2}} \]

### Step 3: Recognize the Riemann sum
As \( n \to \infty \), the expression becomes a Riemann sum for the function:

\[ f(x) = \frac{1}{3 + 8x + 4x^2} \]

where \( x = \frac{r}{n} \). The limit can now be expressed as an integral from 0 to 1:

\[ \int_0^1 \frac{1}{3 + 8x + 4x^2} \, dx \]

### Step 4: Evaluate the integral
To evaluate the integral, we can use partial fraction decomposition or substitution.

1. The denominator can be factored or simplified. The quadratic \( 4x^2 + 8x + 3 \) can be rewritten as:

\[ 4(x + 1)^2 - 1 \]

Thus, we can rewrite the integral as:

\[ \int_0^1 \frac{1}{4(x + 1)^2 - 1} \, dx \]

2. We can use the substitution \( u = 2x + 2 \), which gives \( du = 2dx \) or \( dx = \frac{du}{2} \).

3. The limits change accordingly, and we can evaluate the integral.

### Step 5: Final calculation
After performing the integral and substituting back, we find:

\[ \int_0^1 \frac{1}{3 + 8x + 4x^2} \, dx = \frac{1}{4} \ln \left( \frac{9}{5} \right) \]

### Conclusion
Thus, the final limit is:

\[ \lim_{n \to \infty} n \left( \frac{1}{3n^2 + 8n + 4} + \frac{1}{3n^2 + 16n + 16} + \ldots + \frac{1}{3n^2 + 8n + 4n} \right) = \frac{1}{4} \ln \left( \frac{9}{5} \right) \]