Question

Let f(n) = $\frac{4n + \sqrt{4n^2-1}}{\sqrt{2n+1}+\sqrt{2n-1}}$ and $\lim_{n\to\infty} (\frac{f(1)+f(2)+....f(n)}{\sqrt{1}+\sqrt{2}+\sqrt{3}+....+\sqrt{n}})$ is equal to $\alpha$ then the value of $27\alpha^2$ is:

Answer 1

243/2

Answer 2

Solution Overview:

1. Simplify f(n):
   Write

   $$f(n)=\frac{4n+\sqrt{4n^2-1}}{\sqrt{2n+1}+\sqrt{2n-1}}.$$

   Multiply numerator and denominator by $\sqrt{2n+1}-\sqrt{2n-1}$ noting that $(\sqrt{2n+1}+\sqrt{2n-1})(\sqrt{2n+1}-\sqrt{2n-1})=2$. This gives

   $$f(n)=\frac{1}{2}\Bigl[ \Bigl(4n+\sqrt{4n^2-1}\Bigr)\Bigl(\sqrt{2n+1}-\sqrt{2n-1}\Bigr)\Bigr].$$

2. Express in Telescoping Form:
   Recognize that

   $$\sqrt{4n^2-1}=\sqrt{(2n-1)(2n+1)}.$$

   With some algebraic manipulation, the expression simplifies to

   $$f(n)=\frac{1}{2}\Bigl[(2n+1)\sqrt{2n+1}-(2n-1)\sqrt{2n-1}\Bigr].$$

   This is a telescoping term.

3. Sum the Series:
   Sum $f(n)$ from $n=1$ to $N$:

   $$S_N=\sum_{n=1}^N f(n)=\frac{1}{2}\Bigl[(2N+1)\sqrt{2N+1} - (2\cdot1-1)\sqrt{1}\Bigr]=\frac{1}{2}\Bigl[(2N+1)\sqrt{2N+1}-1\Bigr].$$

4. Estimate the Denominator:
   The denominator $D_N=\sqrt{1}+\sqrt{2}+\cdots+\sqrt{N}$ for large $N$ can be approximated by the integral:

   $$D_N\approx \int_0^N \sqrt{x}\,dx=\frac{2}{3}N^{3/2}.$$

5. Compute the Limit:
   For large $N$,

   $$S_N\sim\frac{1}{2}(2N\cdot\sqrt{2N})=\sqrt{2}\,N^{3/2}.$$

   Therefore,

   $$\alpha=\lim_{N\to\infty}\frac{S_N}{D_N}\sim \frac{\sqrt{2}\,N^{3/2}}{\frac{2}{3}N^{3/2}}=\frac{3\sqrt{2}}{2}.$$

6. Final Calculation:
   Finally, compute

   $$27\alpha^2=27\Bigl(\frac{3\sqrt{2}}{2}\Bigr)^2=27\cdot\frac{9\cdot2}{4}=27\cdot\frac{18}{4}=\frac{27\cdot9}{2}=\frac{243}{2}.$$

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Answer:
$\frac{243}{2}$

Subject: Mathematics (NCERT Class XII, Chapter: Sequences and Series, Topic: Telescoping Series)
Difficulty Level: Medium
Question Type: fill_in_the_blank