Question

The value of $\displaystyle\lim _{n \rightarrow \infty} \frac{1+2-3+4+5-6+\ldots +(3 n-2)+(3 n-1)-3 n}{\sqrt{2 n^4+4 n+3-} \sqrt{n^4+5 n+4}}$ is :

• A. $3(\sqrt{2}+1)$
• B. $\frac{3}{2}(\sqrt{2}+1)$
• C. $\frac{\sqrt{2}+1}{2}$
• D. $\frac{3}{2 \sqrt{2}}$

Answer 1

The Correct Option is B

$\underset{n\to \infty }{\mathrm{Lim}}\frac{0+3+6+9+\dots .n\text{terms}}{\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}}$
$\underset{n\to \infty }{\mathrm{Lim}}\frac{3n(n-1)}{2\left(\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}\right)}$
\(=\frac3{2(\sqrt2-1)}=\frac3{2}(\sqrt2+1)\)

Answer 2

1. The numerator is the sum of a series with alternating signs. Notice the pattern: \[ 1 + 2 - 3, \quad 4 + 5 - 6, \quad 7 + 8 - 9, \ldots \] Each group of three terms sums to \(0\). Therefore, the total sum simplifies to: \[ \text{Sum of the numerator} = \frac{3n}{2}. \] 2. The denominator is the difference of two square roots: \[ \sqrt{2n^4 + 4n + 3} - \sqrt{n^4 + 5n + 4}. \] To simplify, multiply and divide by the conjugate: \[ \sqrt{2n^4 + 4n + 3} - \sqrt{n^4 + 5n + 4} = \frac{(2n^4 + 4n + 3) - (n^4 + 5n + 4)}{\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4}}. \] 3. Simplify the numerator of the fraction: \[ (2n^4 + 4n + 3) - (n^4 + 5n + 4) = n^4 - n - 1. \] 4. Approximate the denominator for large \(n\) by factoring out \(n^4\) from each square root: \[ \sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4} \approx \sqrt{2n^4} + \sqrt{n^4} = n^2(\sqrt{2} + 1). \] 5. Combine results to express the entire limit: \[ \lim_{n \to \infty} \frac{\frac{3n}{2}}{\frac{n^4 - n - 1}{n^2(\sqrt{2} + 1)}} \approx \frac{3}{2}(\sqrt{2} + 1). \] Thus, the value of the limit is: \[ \frac{3}{2}(\sqrt{2} + 1). \] The problem combines the summation of alternating series and the simplification of expressions involving square roots. The numerator simplifies due to periodic cancellation in the series, and the denominator simplifies using the conjugate method.