Problem

Evaluate the Summation sum from n=1 to 6 of (2/3)^(n-1)

The problem is asking you to find the total sum of a series of numbers that follow a specific pattern. The pattern is that each number in the series is obtained by raising (2/3) to a power that is one less than the current term number (n-1), for all term numbers from n=1 to n=6. The task involves calculating each term individually according to this pattern and then finding the total sum of these six terms.

$\sum_{n = 1}^{6} ⁡ \left(\left(\right. \frac{2}{3} \left.\right)\right)^{n - 1}$

Answer

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Solution:

Step 1: Understanding the Geometric Series

To calculate the sum of a finite geometric series, we use the formula $S = a \left(\frac{1 - r^{n}}{1 - r}\right)$, where $S$ is the sum, $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.

Step 2: Identifying the Common Ratio

The common ratio $r$ can be determined by the expression $r = \frac{a_{n+1}}{a_n}$.

Step 2.1: Substituting Terms

We substitute $a_n$ and $a_{n+1}$ into the ratio formula: $r = \frac{(\frac{2}{3})^{(n+1)-1}}{(\frac{2}{3})^{n-1}}$.

Step 2.2: Simplification Process

We proceed to simplify the expression.

Step 2.2.1: Factor Cancellation

We factor out $(\frac{2}{3})^{n-1}$ from $(\frac{2}{3})^{n+1-1}$: $r = \frac{(\frac{2}{3})^{n-1}(\frac{2}{3})^{2-(n-1)}}{(\frac{2}{3})^{n-1}}$.

Step 2.2.2: Simplifying Common Factors

We simplify by multiplying by 1: $r = \frac{(\frac{2}{3})^{n-1}(\frac{2}{3})^{2-(n-1)}}{(\frac{2}{3})^{n-1} \cdot 1}$.

Step 2.2.3: Final Ratio Calculation

We cancel out the common factors and simplify: $r = \frac{(\frac{2}{3})^{2-(n-1)}}{1} = (\frac{2}{3})^{2-(n-1)} = (\frac{2}{3})^{2-n+n-1} = (\frac{2}{3})^{1} = \frac{2}{3}$.

Step 3: Calculating the First Term

We find the first term $a$ by substituting $n=1$ into $(\frac{2}{3})^{n-1}$.

Step 3.1: Substitution for First Term

Substituting $n=1$: $a = (\frac{2}{3})^{1-1}$.

Step 3.2: Simplifying the First Term

We simplify the expression: $a = (\frac{2}{3})^{0} = \frac{2^0}{3^0} = \frac{1}{3^0} = \frac{1}{1} = 1$.

Step 4: Applying the Sum Formula

We substitute the values of $r$, $a$, and the number of terms into the sum formula: $S = 1 \cdot \frac{1 - (\frac{2}{3})^6}{1 - \frac{2}{3}}$.

Step 5: Simplifying the Sum

We simplify the sum expression.

Step 5.1: Initial Simplification

We start by multiplying the numerator and denominator by 3 to clear the fraction: $S = \frac{3(1 - (\frac{2}{3})^6)}{3(1 - \frac{2}{3})}$.

Step 5.2: Distributive Property

We apply the distributive property and cancel common factors: $S = \frac{3 - 3(\frac{2}{3})^6}{3 - 2}$.

Step 5.3: Further Simplification

We continue to simplify the numerator and denominator: $S = \frac{3 - \frac{64}{243}}{1} = \frac{3 \cdot \frac{243}{243} - \frac{64}{243}}{1}$.

Step 5.4: Final Sum Calculation

We combine the fractions and simplify: $S = \frac{729 - 64}{243} = \frac{665}{243}$.

Step 6: Presenting the Result

The sum can be expressed in various forms:

  • Exact Form: $\frac{665}{243}$
  • Decimal Form: $2.73662551\ldots$
  • Mixed Number Form: $2 \frac{179}{243}$

Knowledge Notes:

The problem involves the summation of a finite geometric series. Key concepts include:

  • Geometric Series: A sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the ratio.

  • Sum of Geometric Series: For a finite geometric series, the sum is given by the formula $S = a \left(\frac{1 - r^{n}}{1 - r}\right)$.

  • Common Ratio: The constant factor between consecutive terms of a geometric series, found using $r = \frac{a_{n+1}}{a_n}$.

  • Simplification: The process of reducing expressions to their simplest form, often involving canceling common factors and applying arithmetic operations.

  • Exact, Decimal, and Mixed Number Forms: Different ways to express the result, depending on the desired precision or format.

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