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

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: $2.73662551\dots$
• 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.