Question

MATH HELP QUESTION IN LINK http://prntscr.com/iwuz6h

Answer 1

@iGreen @Vocaloid @563blackghost

Answer 2

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

Which follows the form:

\[\sum_{n = 1}^{\infty} ar^{n - 1} \]

a. Write the first four terms of the series

1.
\[-4(\frac{ 1 }{ 3})^{1 - 1}\]
\[-4(\frac{ 1 }{ 3})^{0}\]
\[-4(1) = -4\]

2.
\[-4(\frac{ 1 }{ 3})^{2 - 1}\]
\[-4(\frac{ 1 }{ 3})^{1}\]
\[-4(\frac{ 1 }{ 3}) = \frac{ -4 }{ 3 }\]

3.
\[-4(\frac{ 1 }{ 3})^{3 - 1}\]
\[-4(\frac{ 1 }{ 3})^{2}\]
\[-4(\frac{ 1^2 }{ 3^2}) \]
\[-4(\frac{ 1 }{ 9}) = \frac{ -4 }{ 9 }\]

4.
\[-4(\frac{ 1 }{ 3})^{4 - 1} \]
\[-4(\frac{ 1 }{ 3})^{3 }\]
\[-4(\frac{ 1^3 }{ 3^3})\]
\[-4(\frac{ 1 }{ 27}) = \frac{ -4 }{ 27 }\]

b. Does the series diverge or converge?

Recall the form,
\[\sum_{n = 1}^{\infty} -4(\frac{ 1 }{ 3})^{n - 1} \]
Where,
\[\sum_{n = 1}^{\infty} ar^{n - 1} \]

A series only converges under the following condition
\[|r| < 1\]

Since the absolute value of r in this case, 1/3, is less than 1, this series converges.

c. If a series has a sum, find the sum

If a series converges, we can then find the sum with the following formula:

\[\frac{ a }{ 1 - r }\]
\[\frac{ -4 }{ 1 - \frac{ 1 }{ 3 } }\]
\[\frac{ -4 }{ \frac{ 2 }{ 3 } }\]
\[-4 \times \frac{ 3 }{ 2 }\]
\[\frac{ -12 }{ 2 } = -6\]

Answer 3

thanks bro