Evaluate the Summation sum from k=1 to infinity of 7(1/3)^(k-1)
The question asks for the evaluation of an infinite series, specifically a summation of terms following a certain pattern. The series starts at k=1 and extends to infinity. Each term in the series is of the form 7(1/3)^(k-1), which suggests it is a geometric series because each term is obtained by multiplying the previous term by a constant common ratio, in this case, 1/3. The problem requires finding the sum of all terms in this series. To solve this problem, one would typically use the formula for the sum of an infinite geometric series, given that the absolute value of the common ratio is less than 1.
$\sum_{k = 1}^{\infty} 7 \left(\left(\right. \frac{1}{3} \left.\right)\right)^{k - 1}$
Answer
Solution:
Step 1:
To determine the sum of an infinite geometric series, use the formula $S = \frac{a}{1 - r}$, where $a$ is the first term and $r$ is the common ratio.
Step 2:
Calculate the common ratio using the formula $r = \frac{a_{k+1}}{a_k}$.
Step 2.1:
Insert $a_k$ and $a_{k+1}$ into the ratio formula: $r = \frac{7(\frac{1}{3})^{(k+1)-1}}{7(\frac{1}{3})^{k-1}}$.
Step 2.2:
Proceed to simplify the expression.
Step 2.2.1:
Eliminate the factor of 7: $r = \frac{\cancel{7}(\frac{1}{3})^{k}}{\cancel{7}(\frac{1}{3})^{k-1}}$.
Step 2.2.2:
Reduce the powers of $\frac{1}{3}$: $r = \frac{(\frac{1}{3})^{k}}{(\frac{1}{3})^{k-1}}$.
Step 2.2.3:
Factor out $(\frac{1}{3})^{k-1}$ from the numerator: $r = \frac{(\frac{1}{3})^{k-1}(\frac{1}{3})^{1}}{(\frac{1}{3})^{k-1}}$.
Step 2.2.4:
Cancel out the common $(\frac{1}{3})^{k-1}$ terms: $r = \frac{\cancel{(\frac{1}{3})^{k-1}}(\frac{1}{3})^{1}}{\cancel{(\frac{1}{3})^{k-1}}}$.
Step 2.2.5:
Simplify to find the common ratio: $r = (\frac{1}{3})^{1}$.
Step 2.2.6:
Conclude that $r = \frac{1}{3}$.
Step 3:
Verify that the series converges by ensuring $|r| < 1$.
Step 4:
Identify the first term of the series by substituting $k = 1$.
Step 4.1:
Calculate the first term: $a = 7(\frac{1}{3})^{1-1}$.
Step 4.2:
Simplify the expression.
Step 4.2.1:
Recognize that any number to the power of 0 is 1: $a = 7(\frac{1}{3})^{0}$.
Step 4.2.2:
Apply this rule: $a = 7 \cdot 1$.
Step 4.2.3:
Conclude that the first term is $a = 7$.
Step 5:
Insert $r$ and $a$ into the sum formula: $S = \frac{7}{1 - \frac{1}{3}}$.
Step 6:
Carry out the simplification.
Step 6.1:
Work on the denominator first.
Step 6.1.1:
Express 1 as a fraction with the same denominator: $S = \frac{7}{\frac{3}{3} - \frac{1}{3}}$.
Step 6.1.2:
Combine the fractions: $S = \frac{7}{\frac{2}{3}}$.
Step 6.2:
Multiply the numerator by the reciprocal of the denominator: $S = 7 \cdot \frac{3}{2}$.
Step 6.3:
Perform the multiplication: $S = \frac{7 \cdot 3}{2}$.
Step 6.4:
Finalize the calculation: $S = \frac{21}{2}$.
Step 7:
Present the sum in various forms.
Exact Form: $\frac{21}{2}$
Decimal Form: $10.5$
Mixed Number Form: $10 \frac{1}{2}$
Knowledge Notes:
Infinite Geometric Series: An infinite geometric series is a sum of the form $a + ar + ar^2 + ar^3 + \ldots$ where $a$ is the first term and $r$ is the common ratio. The series converges to a finite sum if $|r| < 1$.
Convergence Criterion: For an infinite geometric series to converge, the absolute value of the common ratio must be less than 1, i.e., $|r| < 1$.
Sum Formula: The sum of a convergent infinite geometric series is given by $S = \frac{a}{1 - r}$, where $S$ is the sum, $a$ is the first term, and $r$ is the common ratio.
Simplification Techniques: When simplifying expressions involving exponents, common factors can be canceled, and powers can be combined using the laws of exponents.
Fraction Operations: When combining fractions with different denominators, it is often helpful to express them with a common denominator before adding or subtracting. Multiplying by the reciprocal is a method used to divide fractions.
Exponent Rules: Any nonzero number raised to the power of 0 equals 1. This is known as the zero exponent rule.
