How would you write the series 3+12+48+192 using sigma notation?(1 point)

Question

Responses

∑4n=1(3)(4)n−1
sigma-summation underscript n equals 1 overscript 4 End Scripts left parenthesis 3 right parenthesis left parenthesis 4 right parenthesis superscript n minus 1 baseline

∑4n=1(4)(3)n−1
sigma-summation underscript n equals 1 overscript 4 End Scripts left parenthesis 4 right parenthesis left parenthesis 3 right parenthesis superscript n minus 1 baseline

∑4n=1(3)(−4)n−1
sigma-summation underscript n equals 1 overscript 4 End Scripts left parenthesis 3 right parenthesis left parenthesis negative 4 right parenthesis superscript n minus 1 baseline

∑3n=1(3)(4)n−1

Answer 1

To express the series \(3 + 12 + 48 + 192\) in sigma notation, we first recognize that each term can be described in a general form.

The first term is \(3\), the second term is \(12\) (which is \(3 \times 4^1\)), the third term is \(48\) (which is \(3 \times 4^2\)), and the fourth term is \(192\) (which is \(3 \times 4^3\)).

Notice that we can express each term as \(3 \times 4^{n-1}\) where \(n\) starts from 1 to 4.

Thus, the series can be written in sigma notation as:

\[ \sum_{n=1}^{4} 3 \cdot 4^{n-1} \]

From your provided options, this corresponds to:

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

Therefore, the correct response is:

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