Math Series: Sum Of 3(4)^(n-1) From N=1 To 4

Jan 19, 2026 by Andrew McMorgan 45 views

Hey mathletes! Today, we're diving deep into the fascinating world of series, specifically tackling a problem that's all about evaluating a sum and identifying its terms. Get ready to flex those mathematical muscles, guys, because we're going to break down this sigma notation and figure out exactly what's going on. So, let's get started with our series:

\sum_{n=1}^4 3(4)^{n-1}

This notation might look a little intimidating at first glance, but trust me, it's your roadmap to understanding the sequence of numbers we need to add up. The sigma symbol ( $\sum$ ) is the big player here, telling us to sum up a bunch of terms. The 'n=1' at the bottom tells us where to start – we'll begin with n equal to 1. The '4' at the top tells us where to stop – we'll go all the way up to n equal to 4. And the star of the show, the expression $3(4)^{n-1}$ , is what we'll be plugging our 'n' values into to generate each term of the series. Our mission, should we choose to accept it, is twofold: first, to evaluate the entire sum, and second, to correctly identify the individual terms that make up this sum. We'll be looking at four multiple-choice options (A, B, C, and D) to see which one accurately represents the terms of our series. So, buckle up, and let's get this series evaluated!

Understanding the Series Notation

Alright, let's get our heads around this series notation: $\sum_n=1}^4 3(4)^{n-1}$ . Understanding the series notation is absolutely key to solving this problem, so let's break it down piece by piece. The big, Greek letter Sigma ( $\sum$ ) is the universal symbol for summation, meaning we're going to be adding up a sequence of numbers. The subscript, $n=1$ , tells us where our variable, $n$ , begins. In this case, $n$ starts at 1. The superscript, $4$ , is our stopping point. So, $n$ will take on the values 1, 2, 3, and 4. The expression that follows, $3(4)^{n-1}$ , is the formula that generates each term in our series. For every value of $n$ from 1 to 4, we're going to substitute that value into this expression to calculate a specific term. Think of it like a function machine you put in $n$ , and out pops a term of the series. The '3' is a constant multiplier, and the $(4)^{n-1$ part is an exponential term where the base is 4. The exponent, $n-1$ , is crucial because it changes with each value of $n$ . This setup strongly suggests we're dealing with a geometric series, where each term is found by multiplying the previous term by a constant ratio. But before we jump to conclusions, let's do the actual calculation. We need to find the terms for $n=1, n=2, n=3,$ and $n=4$ . Let's plug in each value of $n$ into the formula $3(4)^{n-1}$ .

For n=1: The term is $3(4)^{1-1} = 3(4)^0$ . Remember that any non-zero number raised to the power of 0 is 1. So, this term becomes $3 imes 1 = 3$ . This is our first term.
For n=2: The term is $3(4)^{2-1} = 3(4)^1$ . Any number raised to the power of 1 is itself. So, this term is $3 imes 4 = 12$ . This is our second term.
For n=3: The term is $3(4)^{3-1} = 3(4)^2$ . Here, we need to calculate $4^2$ , which is $4 imes 4 = 16$ . So, the term is $3 imes 16 = 48$ . This is our third term.
For n=4: The term is $3(4)^{4-1} = 3(4)^3$ . First, calculate $4^3$ , which is $4 imes 4 imes 4 = 64$ . So, the term is $3 imes 64 = 192$ . This is our fourth and final term.

So, the individual terms of the series are 3, 12, 48, and 192. Now, let's look at the options provided to see which one matches this sequence.

Identifying the Terms of the Series

Okay, guys, we've done the heavy lifting of calculating each term. Now it's time to identify the terms of the series and match them with the given options. We found that the terms generated by the formula $3(4)^{n-1}$ for $n=1, 2, 3,$ and $4$ are 3, 12, 48, and 192, respectively. Let's scrutinize the options:

Option A: $4+12+48+192$ This option starts with 4. Our first term is definitely 3, not 4. So, this is incorrect.
Option B: $4+12+36$ This option only has three terms, and it starts with 4. Our series has four terms, and the first one is 3. This is incorrect.
Option C: $3+12+48+192$ Let's check this one carefully. It starts with 3, then 12, then 48, and finally 192. This perfectly matches the terms we calculated: 3, 12, 48, 192. Bingo!
Option D: $3+12+48$ This option has only three terms. Our series clearly goes up to $n=4$ , giving us four terms. So, this is incorrect.

Based on our calculations, Option C is the only one that correctly lists all the terms of the series in the correct order. It's super important to pay attention to the starting value of $n$ (which is 1 in this case) and the ending value (which is 4). If the question had started with $n=0$ or ended at $n=3$ , the terms would be different, and so would the correct option. So, always double-check those limits!

Evaluating the Series Sum

We've nailed identifying the terms, but the question also asks us to evaluate the series. This means we need to find the total sum of all the terms we just identified. Our series is $3 + 12 + 48 + 192$ . To evaluate the series, we simply add these numbers together. This is a straightforward addition problem:

3 + 12 + 48 + 192

Let's do this step-by-step:

Add the first two terms: $3 + 12 = 15$
Add the result to the third term: $15 + 48 = 63$
Add the result to the fourth term: $63 + 192 = 255$

So, the sum of the series is 255.

Alternatively, since we recognized this as a geometric series, we could use the formula for the sum of the first $k$ terms of a geometric series, which is $S_k = a rac{r^k - 1}{r - 1}$ , where $a$ is the first term, $r$ is the common ratio, and $k$ is the number of terms.

In our series $3, 12, 48, 192$ :

The first term, $a = 3$ .
To find the common ratio $r$ , we can divide any term by its preceding term: $12/3 = 4$ , $48/12 = 4$ , $192/48 = 4$ . So, the common ratio $r = 4$ .
The number of terms, $k$ , is 4 (since $n$ goes from 1 to 4).

Now, let's plug these values into the formula:

S_4 = 3 rac{4^4 - 1}{4 - 1}

First, calculate $4^4$ : $4^4 = 4 imes 4 imes 4 imes 4 = 256$ .

Now, substitute this back into the formula:

S_4 = 3 rac{256 - 1}{3}

S_4 = 3 rac{255}{3}

We can cancel out the 3s:

S_4 = 255

Both methods give us the same result! This confirms that our evaluation of the series sum is correct. It's always a good idea to have multiple ways to solve a problem, as it helps verify your answer and deepens your understanding.

Final Answer and Key Takeaways

We've successfully navigated through the process of understanding, calculating, and evaluating the given series. The question asks two main things: which option shows the terms of the series, and implicitly, what is the sum of the series. We found that the terms of the series $\sum_{n=1}^4 3(4)^{n-1}$ are $3, 12, 48, 192$ . This corresponds directly to Option C. We also evaluated the sum of the series by both direct addition and using the geometric series sum formula, arriving at a total sum of 255.

So, to recap the main points:

Deconstruct the Sigma Notation: Always break down the summation notation ( $\sum$ ) to understand the starting value of $n$ , the ending value of $n$ , and the expression used to generate each term. This is your blueprint for the series.
Calculate Each Term Systematically: Substitute each value of $n$ within the given range into the expression. Keep your calculations neat and organized to avoid errors.
Match Terms to Options: Carefully compare the calculated terms with the options provided. Pay close attention to the first term and the total number of terms.
Evaluate the Sum: Whether by direct addition or using formulas (like the geometric series sum formula if applicable), calculate the total value of all the terms combined.

Understanding series is a fundamental part of mathematics, and practice makes perfect. Keep working through these problems, and you'll become a series-solving pro in no time! Keep exploring, keep learning, and keep crushing those math problems, guys!