Question

\(\sum_{n=1}^{360}\left(\frac{1}{n \sqrt{n+1}+(n+1) \sqrt{n}}\right)=\) (1) \(\frac{17}{19}\) (2) 2 (3) \(\frac{18}{19}\) (4) \(\frac{20}{19}\) (5) \(\frac{1}{19}\)

Short answer

The answer is \(\frac{18}{19}\).

Step-by-step solution

- Simplify the Expression

Consider the expression inside the summation: \ \[ \frac{1}{n \sqrt{n+1} + (n+1) \sqrt{n}} \] \ We will rationalize the denominator by multiplying the numerator and the denominator by its conjugate: \ \[ \text{Conjugate of } n \sqrt{n+1} + (n+1) \sqrt{n} \text{ is } n \sqrt{n+1} - (n+1) \sqrt{n} \]

- Rationalize the Denominator

Multiply the numerator and denominator by the conjugate: \ \[ \frac{1}{n \sqrt{n+1} + (n+1) \sqrt{n}} \times \frac{n\sqrt{n+1} - (n+1)\sqrt{n}}{n\sqrt{n+1} - (n+1)\sqrt{n}} = \frac{n \sqrt{n+1} - (n+1) \sqrt{n}}{(n\sqrt{n+1})^2 - ((n+1)\sqrt{n})^2} \] \ Simplify the denominator: \ \[ (n\sqrt{n+1})^2 = n^2(n+1) \] \ and \ \[ ((n+1)\sqrt{n})^2 = (n+1)^2 n \] \ So the denominator becomes: \ \[ n^2(n+1) - n(n+1)^2 \]

- Simplify Further

Further simplifying the denominator expression: \ \[ n^2(n+1) - n(n+1)^2 = n^2(n+1) - n(n^2 + 2n + 1) = n^3 + n^2 - n^3 - 2n^2 - n = -n^2 - n \] \ Thus, \ \[ \frac{n \sqrt{n+1} - (n+1) \sqrt{n}}{-n^2 - n} = \frac{n \sqrt{n+1} - (n+1) \sqrt{n}}{-n(n+1)} \]

- Separate into Two Fractions

Separate the expression into two fractions: \ \[ \frac{n \sqrt{n+1}}{-n(n+1)} - \frac{(n+1) \sqrt{n}}{-n(n+1)} \] \ Simplify each fraction: \ \[ \frac{\sqrt{n+1}}{-(n+1)} - \frac{\sqrt{n}}{-n} = -\frac{\sqrt{n+1}}{n+1} + \frac{\sqrt{n}}{n} \]

- Summing the Series

Rewrite the simplified expression for the summation: \ \[ \sum_{n=1}^{360} \left( -\frac{\sqrt{n+1}}{n+1} + \frac{\sqrt{n}}{n} \right) \] \ This expression is a telescoping series, where most terms will cancel out. The remaining terms will be: \ \[ - \frac{\sqrt{361}}{361} + 1 = -\frac{19}{361} + 1 \] \ Simplify the final expression: \ \[ 1 - \frac{19}{361} = \frac{361-19}{361} = \frac{342}{361} = \frac{18}{19} \]

Key concepts

Summation

Summation is a mathematical operation that adds together a sequence of numbers. In notation, it is represented by the Greek letter Sigma (\( \Sigma \)). For example, \( \sum_{n=1}^{360} a_n \) refers to the sum of the sequence \( a_n \) from \( n = 1 \) to \( n = 360 \). This concept is critical in many scientific and engineering disciplines.
In the given exercise, summation helps in finding the combined value of expressions over a defined range.
Key points to note are:

• Identify the sequence being summed.
• Determine the range over which the summation occurs.
• Apply properties of summation when necessary.

Sometimes, simplifying the expression inside the summation can make it easier to find the total sum.
As seen in our example, simplifying the complex fraction made the series manageable for summation.

Telescoping Series

A telescoping series is a series where each term cancels out a part of the previous or subsequent term. This makes it easier to find the sum of the series.
Telescoping series can be identified by their form and by trying to write each term in a way that reveals this cancellation property.
Here are the steps to work with telescoping series:

• Rewrite the series to reveal its telescoping nature.
• Identify the terms that cancel each other.
• Sum the remaining terms, which typically are the first and last terms of the series.

In our exercise, after simplifying, we get expressions like \( -\frac{\sqrt{n+1}}{n+1} \) and \( \frac{\sqrt{n}}{n} \) which cancel out most of the terms when summed over the range from 1 to 360.
This results in only the first and last few terms remaining, simplifying the process significantly.

Rationalization

Rationalization is the process of eliminating the square root or cube root from a denominator by multiplying the numerator and denominator by a conjugate or some similar term.
The conjugate of a binomial like \( a + b \sqrt{d} \) is \( a - b \sqrt{d} \). By multiplying by the conjugate, the denominators become rational numbers.
Steps to rationalize:

• Identify the conjugate.
• Multiply both numerator and denominator by this conjugate.
• Simplify both numerator and denominator to find a more straightforward expression.

In the given problem, we rationalized the expression \( \frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}} \) by multiplying with its conjugate to simplify the structure. This allowed complex terms to be managed more easily, ultimately transforming them into an easier series for summation.
This simplification transforms the complex fraction into one that is suitable for further steps towards solving the summation problem.