Question

HELP?!

Answer 1

with what

Answer 2

sorry cant help :(

Answer 3

For part a

The first term is plugging in 1 into the equation and finding the answer
For the second term, plug in 2

And so on

Answer 4

Okay what are the numbers we're using? @FortyTheRapper

Answer 5

we can rewrite the series, like below:
\[ \Large - 4{\sum\limits_{n = 1}^\infty {\left( {\frac{1}{3}} \right)} ^n} \cdot {\left( {\frac{1}{3}} \right)^{ - 1}} = - 12{\sum\limits_{n = 1}^\infty {\left( {\frac{1}{3}} \right)} ^n}\]

Answer 6

then we have:
\[\Large \begin{gathered}
- 4{\sum\limits_{n = 1}^\infty {\left( {\frac{1}{3}} \right)} ^n} \cdot {\left( {\frac{1}{3}} \right)^{ - 1}} = - 12{\sum\limits_{n = 1}^\infty {\left( {\frac{1}{3}} \right)} ^n} = \hfill \\
\hfill \\
= - 12\left( {\frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}}} \right) - 12{\sum\limits_{n = 5}^\infty {\left( {\frac{1}{3}} \right)} ^n} = \hfill \\
\hfill \\
= \left( { - 4 + \frac{{ - 4}}{3} + \frac{{ - 4}}{9} + \frac{{ - 4}}{{27}}} \right) - 12{\sum\limits_{n = 5}^\infty {\left( {\frac{1}{3}} \right)} ^n} \hfill \\
\end{gathered} \]

Answer 7

-4/1, -4/3, -4/9, -4/27?? Is what we get for part a?

Answer 8

that's right!

Answer 9

And part b how do we know?!

Answer 10

furthermore, please note that, the series:
\[\huge{\sum\limits_{n = 1}^\infty {\left( {\frac{1}{3}} \right)} ^n}\]
is a series with all positive terms, then in order to establish if it converges or diverges, we can apply the ratio criterion, for example
what can you conclude?

Answer 11

It gets bigger!

Answer 12

are you sure? It is the geometrical series

Answer 13

Okay tell me.

Answer 14

such series is a convergent series, here is why:
\[\huge \begin{gathered}
\sum\limits_{n = 1}^\infty {{a_n}} = {\sum\limits_{n = 1}^\infty {\left( {\frac{1}{3}} \right)} ^n} \hfill \\
\hfill \\
\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{3^n}}}{{{3^{n + 1}}}} = \frac{1}{3} < 1 \hfill \\
\end{gathered} \]

Answer 15

3^{n-1} so its convergent.

Answer 16

Please it is convergent, since the criterion of ratio gives a quantity less than \(1\)
and we can write:
\[\Large \sum\limits_{n = 1}^\infty {{a_n}} = {\sum\limits_{n = 1}^\infty {\left( {\frac{1}{3}} \right)} ^n} = \frac{1}{3} \cdot \frac{1}{{1 - \frac{1}{3}}} = ...?\]

Answer 17

so the requested sum, is:
\[ \large - 4{\sum\limits_{n = 1}^\infty {\left( {\frac{1}{3}} \right)} ^n} \cdot {\left( {\frac{1}{3}} \right)^{ - 1}} = - 12{\sum\limits_{n = 1}^\infty {\left( {\frac{1}{3}} \right)} ^n} = - 12 \cdot \frac{1}{3} \cdot \frac{1}{{1 - \frac{1}{3}}} = ...?\]

Answer 18

\[\Large \begin{gathered}
- 4{\sum\limits_{n = 1}^\infty {\left( {\frac{1}{3}} \right)} ^n} \cdot {\left( {\frac{1}{3}} \right)^{ - 1}} = - 12{\sum\limits_{n = 1}^\infty {\left( {\frac{1}{3}} \right)} ^n} = \hfill \\
\hfill \\
= - 12 \cdot \frac{1}{3} \cdot \frac{1}{{1 - \frac{1}{3}}} = ...? \hfill \\
\end{gathered} \]

Answer 19

Wait that whole thing is teh sum?

Answer 20

yes!

Answer 21

please what is:
\[ \huge - 12 \cdot \frac{1}{3} \cdot \frac{1}{{1 - \frac{1}{3}}} = ...?\]

Answer 22

\[-4*\frac{1}{1-1/3}\]
\[-4*\frac{1}{2/3}\]
\[-4*\frac{3}{2}=-6\]

Answer 23

that's right!

Answer 24

Wow thank you!

Answer 25

:)