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.
$\sum_{n = 1}^{6} \left(\left(\right. \frac{2}{3} \left.\right)\right)^{n - 1}$
Solution:
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.
The common ratio $r$ can be determined by the expression $r = \frac{a_{n+1}}{a_n}$.
We substitute $a_n$ and $a_{n+1}$ into the ratio formula: $r = \frac{(\frac{2}{3})^{(n+1)-1}}{(\frac{2}{3})^{n-1}}$.
We proceed to simplify the expression.
We factor out $(\frac{2}{3})^{n-1}$ from $(\frac{2}{3})^{n+1-1}$: $r = \frac{(\frac{2}{3})^{n-1}(\frac{2}{3})^{2-(n-1)}}{(\frac{2}{3})^{n-1}}$.
We simplify by multiplying by 1: $r = \frac{(\frac{2}{3})^{n-1}(\frac{2}{3})^{2-(n-1)}}{(\frac{2}{3})^{n-1} \cdot 1}$.
We cancel out the common factors and simplify: $r = \frac{(\frac{2}{3})^{2-(n-1)}}{1} = (\frac{2}{3})^{2-(n-1)} = (\frac{2}{3})^{2-n+n-1} = (\frac{2}{3})^{1} = \frac{2}{3}$.
We find the first term $a$ by substituting $n=1$ into $(\frac{2}{3})^{n-1}$.
Substituting $n=1$: $a = (\frac{2}{3})^{1-1}$.
We simplify the expression: $a = (\frac{2}{3})^{0} = \frac{2^0}{3^0} = \frac{1}{3^0} = \frac{1}{1} = 1$.
We substitute the values of $r$, $a$, and the number of terms into the sum formula: $S = 1 \cdot \frac{1 - (\frac{2}{3})^6}{1 - \frac{2}{3}}$.
We simplify the sum expression.
We start by multiplying the numerator and denominator by 3 to clear the fraction: $S = \frac{3(1 - (\frac{2}{3})^6)}{3(1 - \frac{2}{3})}$.
We apply the distributive property and cancel common factors: $S = \frac{3 - 3(\frac{2}{3})^6}{3 - 2}$.
We continue to simplify the numerator and denominator: $S = \frac{3 - \frac{64}{243}}{1} = \frac{3 \cdot \frac{243}{243} - \frac{64}{243}}{1}$.
We combine the fractions and simplify: $S = \frac{729 - 64}{243} = \frac{665}{243}$.
The sum can be expressed in various forms:
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.