Key Concepts and Formulas:

1. General Term of a Sequence: Identifying a formula for the $r$-th term ($a_r$) of a sequence is crucial for summation. This often involves recognizing patterns in numerators, denominators, or differences between terms, frequently related to arithmetic or geometric progressions.
2. Summation of Series: The ability to express and evaluate sums of sequences is fundamental. For a geometric progression (GP) with first term $A$, common ratio $R$, and $N$ terms, the sum is $S_N = A \frac{1-R^N}{1-R}$ when $|R| < 1$.
3. Properties of Summation: The linearity of summation allows splitting a sum of terms into sums of individual components: $\sum (x_r + y_r) = \sum x_r + \sum y_r$ and $\sum c \cdot x_r = c \sum x_r$. Also, $\sum_{r=1}^N c = Nc$.
4. Floor Function ($\lfloor x \rfloor$): This function returns the greatest integer less than or equal to $x$. It effectively "rounds down" to the nearest integer.

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Step-by-Step Solution:

Step 1: Identify the General Term ($a_r$) of the Sequence

We are given the sequence: $\frac{1}{3}, \frac{5}{9}, \frac{19}{27}, \frac{65}{81}, \dots$ Let's analyze the structure of each term.

• Denominators: The denominators are $3, 9, 27, 81, \dots$. These are powers of 3: $3^1, 3^2, 3^3, 3^4, \dots$. Thus, the denominator of the $r$-th term is $3^r$.

• Numerators: The numerators are $1, 5, 19, 65, \dots$. Let's see if there's a relationship with the corresponding denominators ($3^r$).

  • For $r=1$: Denominator is $3^1=3$. Numerator is $1$. We can write $1 = 3 - 2$.
  • For $r=2$: Denominator is $3^2=9$. Numerator is $5$. We can write $5 = 9 - 4$.
  • For $r=3$: Denominator is $3^3=27$. Numerator is $19$. We can write $19 = 27 - 8$.
  • For $r=4$: Denominator is $3^4=81$. Numerator is $65$. We can write $65 = 81 - 16$.

  The numbers being subtracted are $2, 4, 8, 16, \dots$, which are powers of 2: $2^1, 2^2, 2^3, 2^4, \dots$. So, the number subtracted for the $r$-th term is $2^r$. Therefore, the numerator of the $r$-th term is $3^r - 2^r$.

• General Term: Combining the numerator and denominator, the $r$-th term $a_r$ is: $a_r = \frac{3^r - 2^r}{3^r}$ To simplify this for summation, we can split the fraction: $a_r = \frac{3^r}{3^r} - \frac{2^r}{3^r} = 1 - \left(\frac{2}{3}\right)^r$

Step 2: Express the Sum of the First 100 Terms ($S_{100}$)

We need to calculate the sum of the first 100 terms, $S_{100}$. Using the general term: $S_{100} = \sum_{r=1}^{100} a_r = \sum_{r=1}^{100} \left(1 - \left(\frac{2}{3}\right)^r\right)$

Step 3: Evaluate the Summation

Using the linearity property of summation, we can split the sum into two parts: $S_{100} = \sum_{r=1}^{100} 1 - \sum_{r=1}^{100} \left(\frac{2}{3}\right)^r$

• Part 1: Sum of the constant term The sum of the constant 1, 100 times, is: $\sum_{r=1}^{100} 1 = 1 \times 100 = 100$

• Part 2: Sum of the geometric progression The second part is the sum of a geometric progression: $\sum_{r=1}^{100} \left(\frac{2}{3}\right)^r$. Here, the first term is $A = (\frac{2}{3})^1 = \frac{2}{3}$. The common ratio is $R = \frac{2}{3}$. The number of terms is $N = 100$. Since $|R| = |\frac{2}{3}| < 1$, we use the GP sum formula $S_N = A \frac{1-R^N}{1-R}$: $\sum_{r=1}^{100} \left(\frac{2}{3}\right)^r = \frac{\frac{2}{3} \left(1 - \left(\frac{2}{3}\right)^{100}\right)}{1 - \frac{2}{3}}$ The denominator is $1 - \frac{2}{3} = \frac{1}{3}$. $= \frac{\frac{2}{3} \left(1 - \left(\frac{2}{3}\right)^{100}\right)}{\frac{1}{3}} = 2 \left(1 - \left(\frac{2}{3}\right)^{100}\right) = 2 - 2\left(\frac{2}{3}\right)^{100}$

Step 4: Combine the Parts to Find the Total Sum $S_{100}$

Substitute the results from Step 3 back into the expression for $S_{100}$: $S_{100} = 100 - \left(2 - 2\left(\frac{2}{3}\right)^{100}\right)$ $S_{100} = 100 - 2 + 2\left(\frac{2}{3}\right)^{100}$ $S_{100} = 98 + 2\left(\frac{2}{3}\right)^{100}$

Step 5: Apply the Floor Function

We need to find the greatest integer less than or equal to $S_{100}$, denoted by $\lfloor S_{100} \rfloor$. The sum is $S_{100} = 98 + 2\left(\frac{2}{3}\right)^{100}$.

Let's analyze the term $2\left(\frac{2}{3}\right)^{100}$:

• The base $\frac{2}{3}$ is a positive number between 0 and 1.
• When a number between 0 and 1 is raised to a large positive power (like 100), the result is a very small positive number that approaches 0.
• Therefore, $0 < \left(\frac{2}{3}\right)^{100} < 1$.
• Multiplying by 2, we get $0 < 2\left(\frac{2}{3}\right)^{100} < 2$.
• Crucially, $2\left(\frac{2}{3}\right)^{100}$ is a tiny positive decimal fraction, significantly less than 1.

So, $S_{100} = 98 + (\text{a very small positive fraction})$. This means $S_{100}$ is slightly greater than 98 but less than 99. For instance, it might be $98.0000...001$.

Applying the floor function: $\lfloor S_{100} \rfloor = \left\lfloor 98 + 2\left(\frac{2}{3}\right)^{100} \right\rfloor = 98$

The greatest integer less than or equal to the sum of the first 100 terms is 98.

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Common Mistakes & Tips:

• Pattern Recognition: Be systematic when finding the general term. Test your derived formula with the initial terms of the sequence.
• Algebraic Simplification: Simplifying the general term by splitting fractions can significantly ease the summation process.
• GP Formula Application: Ensure you correctly identify the first term, common ratio, and number of terms for the GP. Be mindful of the condition $|R| < 1$ for the formula used.
• Floor Function Interpretation: Remember that the floor function always rounds down to the nearest integer, even for numbers just above an integer.

Summary:

The problem involves finding the general term of a sequence by recognizing patterns in its numerators and denominators. The general term was found to be $a_r = 1 - (\frac{2}{3})^r$. The sum of the first 100 terms was then calculated by splitting this into the sum of a constant and the sum of a geometric progression. The sum was found to be $98 + 2(\frac{2}{3})^{100}$. Since $2(\frac{2}{3})^{100}$ is a very small positive quantity, the greatest integer less than or equal to this sum is 98.

The final answer is $\boxed{98}$.