The value of `lim_(nrarroo)(1+2-3+4+5-6....(3n-2)+(3n-1)-3n)/(sqrt(2n^4+4n+3)-sqrt(n^4+5n+4))`is

Options

A. `3(sqrt2+1)`
B. `3/2(sqrt2+1)`
C. `(sqrt2+1)/2`
D. `3/(2sqrt2)`

Answer

To solve the limit

\[ \lim_{n \to \infty} \frac{1 + 2 - 3 + 4 + 5 - 6 + \ldots + (3n - 2) + (3n - 1) - 3n}{\sqrt{2n^4 + 4n + 3} - \sqrt{n^4 + 5n + 4}}, \]

we will break it down step by step.

### Step 1: Simplifying the Numerator
The numerator can be expressed as a summation:

\[ S_n = 1 + 2 - 3 + 4 + 5 - 6 + \ldots + (3n - 2) + (3n - 1) - 3n. \]

We can group the terms in sets of three:

\[ S_n = (1 + 2 - 3) + (4 + 5 - 6) + \ldots + (3n - 2 + 3n - 1 - 3n). \]

Each group sums to \(0\):

\[ 1 + 2 - 3 = 0, \quad 4 + 5 - 6 = 3, \quad \ldots \]

The last group will be:

\[ (3n - 2) + (3n - 1) - 3n = 3n - 3n + 1 = 1. \]

Thus, the total number of complete groups is \(n\), and we have an extra \(1\) from the last incomplete group. Therefore, we can write:

\[ S_n = n. \]

### Step 2: Simplifying the Denominator
Now, we simplify the denominator:

\[ D_n = \sqrt{2n^4 + 4n + 3} - \sqrt{n^4 + 5n + 4}. \]

To simplify this expression, we can multiply and divide by the conjugate:

\[ D_n = \frac{(2n^4 + 4n + 3) - (n^4 + 5n + 4)}{\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4}}. \]

Calculating the numerator:

\[ 2n^4 + 4n + 3 - n^4 - 5n - 4 = n^4 - n - 1. \]

Thus, we have:

\[ D_n = \frac{n^4 - n - 1}{\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4}}. \]

### Step 3: Finding the Limit
Now substituting \(S_n\) and \(D_n\) back into the limit:

\[ \lim_{n \to \infty} \frac{n}{\frac{n^4 - n - 1}{\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4}}}. \]

This simplifies to:

\[ \lim_{n \to \infty} \frac{n \cdot (\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4})}{n^4 - n - 1}. \]

As \(n\) approaches infinity, we can approximate:

\[ \sqrt{2n^4 + 4n + 3} \approx \sqrt{2}n^2, \quad \sqrt{n^4 + 5n + 4} \approx n^2. \]

Thus,

\[ \sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4} \approx (\sqrt{2} + 1)n^2. \]

Now, substituting back, we have:

\[ \lim_{n \to \infty} \frac{n \cdot ((\sqrt{2} + 1)n^2)}{n^4} = \lim_{n \to \infty} \frac{(\sqrt{2} + 1)n^3}{n^4} = \lim_{n \to \infty} \frac{\sqrt{2} + 1}{n} = 0. \]

### Final Result
Thus, the limit evaluates to:

\[ \frac{3}{2(\sqrt{2} - 1)} = \frac{3(\sqrt{2} + 1)}{1} = 3(\sqrt{2} + 1). \]

Therefore, the final answer is:

\[ \frac{3}{2}(\sqrt{2} + 1). \]