Problem 55 - Algebra and Trigonometry

Question

Expand each sum. \(\sum_{k=0}^{n-1} \frac{1}{3^{k+1}}\)

Step-by-Step Solution

Answer

The expanded sum is \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots + \frac{1}{3^n}\).

Step 1 Title - Understand the Summation

The given summation is \(\sum_{k=0}^{n-1} \frac{1}{3^{k+1}}\). This indicates that we need to calculate the sum of the terms \(\frac{1}{3^{k+1}}\) for values of \(k\) starting from \(0\) and going up to \((n-1)\).

Step 2 Title - Write Out the First Few Terms

Substitute \(k = 0, 1, 2, \ldots, n-1\) into the term \(\frac{1}{3^{k+1}}\). The first few terms are: \(\frac{1}{3^1}, \frac{1}{3^2}, \frac{1}{3^3},\ \ldots,\ \frac{1}{3^n}\).

Step 3 Title - Expand the Summation

Expand the summation notation into an explicit sum: \(\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots + \frac{1}{3^n}\).

Step 4 Title - Conclusion

The expanded form of the summation \(\sum_{k=0}^{n-1} \frac{1}{3^{k+1}}\) is \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots + \frac{1}{3^n}\).

Key Concepts

geometric series

A geometric series is a series with a constant ratio between successive terms. In the given exercise, the ratio between each term is \(\frac{1}{3}\). Here's how you can recognize a geometric series:

• Each term is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio.
• In our sum \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots + \frac{1}{3^n}\), the common ratio \(\frac{1}{3}\). So, each term is \(\frac{1}{3}\) times the previous term.

To write it out more formally, if \{a_1, a_2, a_3, \ldots\}\ is a geometric series, then:
\{a, ar, ar^2, ar^3, \ldots\}\ where 'a' is the first term and 'r' is the common ratio.

series expansion

Series expansion involves writing out the terms of a series to make summing them up easier. For the sum \(\frac{1}{3^1} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots + \frac{1}{3^n}\), you deal with each term individually.
To expand it, we substitute each value of k from 0 to (n-1) into \(\frac{1}{3^{k+1}}\):

• For k=0: \(\frac{1}{3^{0+1}} = \frac{1}{3}\).
• For k=1: \(\frac{1}{3^{1+1}} = \frac{1}{9}\).
• For k=2: \(\frac{1}{3^{2+1}} = \frac{1}{27}\).

This process continues until the (n-1) term \(\frac{1}{3^n}\). This step-by-step expansion helps you sum the series more easily.

algebraic notation

Algebraic notation helps in representing series in a concise form using symbols and variables. In the initial exercise, the series is written as: \(\sum_{k=0}^{n-1} \frac{1}{3^{k+1}}\).
Here's how the algebraic notation works:

• The summation symbol \(\sum\) indicates that you are summing up terms.
• The variable 'k' is called the index of summation, starting at 0 and ending at \((n-1)\).
• The expression \(\frac{1}{3^{k+1}}\) is the general term being summed.

Algebraic notation makes it easier to write and manipulate series, particularly when dealing with long or complex sums.

finite series

A finite series is a series that has a definite number of terms. Unlike an infinite series that goes on forever, a finite series stops after a certain number of terms.
In the exercise, the series is finite because it stops at the term \(\frac{1}{3^n}\).

• Finite series are simpler to sum because of their limited number of terms.
• You can use specific formulas to easily sum them.

For a finite geometric series of the form \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots + \frac{1}{3^n}\), the sum is given by: \(\frac{a(1-r^n)}{1-r}\), where 'a' is the first term and 'r' is the common ratio. Using this, you can quickly find the sum rather than adding each term individually.