Express the sum in sigma notation.
-$$\frac { 1 } { 3 } + \frac { 1 } { 9 } + \frac { 1 } { 27 } + \frac { 1 } { 81 }$$
A) $\sum _ { \mathrm { k } = 0 } ^ { 4 } \left( \frac { 1 } { 3 } \right) ^ { \mathrm { k } + 1 }$
B) $\sum _ { \mathrm { k } = 1 } ^ { 4 } \left( \frac { 1 } { 3 } \right) ^ { \mathrm { k } - 1 }$
C) $\sum _ { k = 1 } ^ { 5 } \left( \frac { 1 } { 3 } \right) ^ { \mathrm { k } - 1 }$
D) $\sum _ { \mathrm { k } = 1 } ^ { 4 } \left( \frac { 1 } { 3 } \right) ^ { \mathrm { k } }$

Question

Quizplus · Accepted Answer

In sigma notation, the general form of a geometric series would be $\sum_{k=0}^{n}ar^k$, where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.

To express the given sum in sigma notation, we first notice that the common ratio is $\frac13$. The first term isn't immediately obvious, but we can write:

$$\begin{aligned} \frac13+\frac19+\frac{1}{27}+\frac{1}{81} &= \frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\ &=\frac{1}{3^0}\cdot\frac{1}{3^1}+\frac{1}{3^1}\cdot\frac{1}{3^1}+\frac{1}{3^2}\cdot\frac{1}{3^1}+\frac{1}{3^3}\cdot\frac{1}{3^1}\ &=\sum_{k=1}^{4}\frac{1}{3^k}\cdot\frac13\ &=\sum_{k=1}^{4}\frac{1}{3^{k+1}} \end{aligned}$$

Therefore, the sum can be expressed as $\boxed{\textbf{(D) } \sum_{k=1}^{4}\left(\frac13\right)^k }$.

Express the Sum in Sigma Notation 13+19+127+181\frac { 1 } { 3 } + \frac { 1 } { 9 } + \frac { 1 } { 27 } + \frac { 1 } { 81 }31​+91​+271​+811​

Express the Sum in Sigma Notation $\frac { 1 } { 3 } + \frac { 1 } { 9 } + \frac { 1 } { 27 } + \frac { 1 } { 81 }$