Evaluate the Summation sum from k=1 to infinity of 1/6*7^(k-1)
The question is asking for the evaluation of an infinite series. Specifically, it requests the summation of the terms 1/(6*7^(k-1)) for each integer value of k starting from 1 and continuing indefinitely (k = 1, 2, 3, ...). In the series, each term is a fraction where the numerator is 1 and the denominator is the product of 6 and 7 raised to the power of (k-1). The task is to determine the sum of all such terms, essentially finding the value to which this series converges, if it does indeed converge to a specific value.
$\sum_{k = 1}^{\infty} \frac{1}{6} \cdot 7^{k - 1}$
Answer
Solution:
Step 1:
To determine the sum of an infinite geometric series, we use the sum formula $S = \frac{a}{1 - r}$, where $a$ is the first term and $r$ is the common ratio.
Step 2:
Calculate the common ratio $r$ by using the formula $r = \frac{a_{k+1}}{a_k}$ and performing algebraic simplification.
Step 2.1:
Insert $a_k$ and $a_{k+1}$ into the ratio formula: $r = \frac{\frac{1}{6} \cdot 7^{(k+1)-1}}{\frac{1}{6} \cdot 7^{k-1}}$.
Step 2.2:
Proceed to simplify the expression.
Step 2.2.1:
Eliminate the shared $\frac{1}{6}$ term.
Step 2.2.1.1:
After removing the common factor: $r = \frac{\cancel{\frac{1}{6}} \cdot 7^{k}}{\cancel{\frac{1}{6}} \cdot 7^{k-1}}$.
Step 2.2.1.2:
Reformulate the expression: $r = \frac{7^k}{7^{k-1}}$.
Step 2.2.2:
Cancel out the common $7^{k-1}$ term.
Step 2.2.2.1:
Factor out $7^{k-1}$ from $7^k$: $r = \frac{7^{k-1} \cdot 7^{1}}{7^{k-1}}$.
Step 2.2.2.2:
Eliminate the common factors.
Step 2.2.2.2.1:
Introduce multiplication by $1$: $r = \frac{7^{k-1} \cdot 7^{1}}{7^{k-1} \cdot 1}$.
Step 2.2.2.2.2:
Remove the common factor: $r = \frac{\cancel{7^{k-1}} \cdot 7^{1}}{\cancel{7^{k-1}} \cdot 1}$.
Step 2.2.2.2.3:
Rephrase the expression: $r = \frac{7^{1}}{1}$.
Step 2.2.2.2.4:
Divide $7^1$ by $1$: $r = 7^1$.
Step 2.2.3:
Combine $k$ and $0$: $r = 7^{1}$.
Step 2.2.4:
Simplify each term.
Step 2.2.4.1:
Apply the distributive property: $r = 7^{1}$.
Step 2.2.4.2:
Multiply $-1$ by $-1$: $r = 7^{1}$.
Step 2.2.5:
Subtract $k$ from $k$: $r = 7^{1}$.
Step 2.2.6:
Add $0$ and $1$: $r = 7^{1}$.
Step 2.2.7:
Evaluate the exponent: $r = 7$.
Step 3:
Verify if the series converges or diverges. Given that $|r| \geq 1$, the series is divergent.
Knowledge Notes:
An infinite geometric series is a series with a constant ratio between successive terms. It can be expressed as $a + ar + ar^2 + ar^3 + \ldots$ where $a$ is the first term and $r$ is the common ratio.
The sum of an infinite geometric series converges to a finite value if the absolute value of the common ratio $|r|$ is less than 1. The formula for the sum is $S = \frac{a}{1 - r}$.
If $|r|$ is equal to or greater than 1, the series does not converge to a finite sum and is considered divergent.
Simplifying expressions and canceling common factors are essential algebraic techniques used in evaluating series and other mathematical expressions.
The distributive property is a fundamental algebraic property which states that $a(b + c) = ab + ac$.
Exponents represent repeated multiplication, and understanding the laws of exponents is crucial for simplifying expressions involving powers.
