The problem requires finding the sum of the expression: \(\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \dots + \frac{1}{\sqrt{80} + \sqrt{81}}\).
To tackle this, we will use the concept of rationalizing the denominator.
Step 1: Rationalize Each Term
To rationalize the denominator \(\sqrt{n} + \sqrt{n+1}\), we multiply the numerator and the denominator by the conjugate, \(\sqrt{n+1} - \sqrt{n}\):
\(\frac{1}{\sqrt{n} + \sqrt{n+1}} = \frac{\sqrt{n+1} - \sqrt{n}}{(\sqrt{n} + \sqrt{n+1})(\sqrt{n+1} - \sqrt{n})}\)
The denominator becomes:
\((\sqrt{n+1})^2 - (\sqrt{n})^2 = n+1 - n = 1\)
Thus, the expression simplifies to:
\(\sqrt{n+1} - \sqrt{n}\)
Step 2: Apply to the Sum
Substitute back into the given sum:
\(\sum_{n=1}^{80} \left( \sqrt{n+1} - \sqrt{n} \right)\)
This is a telescoping series, where consecutive terms cancel each other:
So, \((\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \dots + (\sqrt{81} - \sqrt{80})\)
Most terms cancel except the very first negative and the very last positive:
Thus, we are left with:
\(\sqrt{81} - \sqrt{1} = 9 - 1 = 8\)
Conclusion
The sum of the expression is \(8\).
Therefore, the correct answer is: 8.
Important Questions from Series
- In the sequence 6, 9, 14, $x$, 30, 41, a possible value of $x$ is
Let $a_0 = 0$ and define $a_n = \frac{1}{2}(1 + a_{n-1})$ for all positive integers $n \ge 1$.
The least value of $n$ for which $|1 - a_n| < \frac{1}{2^{10}}$ is __________.
(Answer in integer)
- Calculate the reciprocal of the coefficient of $z^3$ in the Taylor series expansion of the function $f(z) = \sin(z)$ around $z = 0$. (Provide the answer as an integer.)
- Let $a_1 = 1$ and $a_n = a_{n-1} + 4$, $n \ge 2$. Then,
$\lim_{n\to\infty} \left[\frac{1}{a_1a_2} + \frac{1}{a_2a_3} + \dots + \frac{1}{a_{n-1}a_n}\right]$
is equal to ________ - Let $S(x) = a_0 + \sum_{n=1}^\infty(a_n \cos (n x) + b_n \sin (n x))$ be the Fourier series of the$2 \pi$ periodic function defined by $f(x) = x^2 + 4 \sin (x) \cos(x)$, $-\pi \le x \le \pi$. Then
$|\sum_{n=0}^\infty a_n - \sum_{n=1}^\infty b_n|$
is equal to ________
