Question

Infinite Geometric Series Sigma Problem will be posted below....I don't know how to find r...

Answer 1

\[\sum_{n=1}^{\infty} (2/3)(-3/4)^{n-1}\]

Answer 2

\[\sum_{n=1}^{\infty} \frac {2}{3} \times (\frac {-3}{4})^{n-1} = \sum_{n = 1}^{\infty} \frac {2}{3} \times \frac {-4}{3} \times = \frac {8}{9} \sum_{n=1}^{\infty} (\frac {-3}{4})^{n} = \frac {8}{9} \times \frac {\frac {-3}{4}}{1 - \frac {-3}{4}} = \frac {8}{9} \times \frac {-3}{7} = \frac {-8}{21}\]

Answer 3

= -8/21

Answer 4

does \[a _{1}\] equal 2/3?

Answer 5

because i thought you inserted the n=1 into the exponent of n-1? or am i wrong?

Answer 6

The answer should be \[ \frac 8 {21}\]
Note that
\[ \sum_{n=1}^\infty a^{n-1} = \frac 1{1-a}, \quad |a| <1
\]
Apply this with
\[ a= -\frac 3 4
\]
to your series and multiply what you get by
\[ \frac 23\]
to obtain
\[
\frac 1 {28} \]

Answer 7

so is the number before the exponent always r?

Answer 8

Yes, but you should examine the first few terms. For example the following sum
\[ \sum_{n=2}^\infty a^{n-1} = a + a^2 +a^3 +\cdots =\frac a{1-a},
\quad |a| <1\\
a \sum_{n=1}^\infty a^{n-1} = a \left (1+ a + a^2 +a^3 +\cdots\right) =a \left(\frac1{1-a}\right)=\frac a{1-a},
\quad |a| <1
\]

Answer 9

I have no idea what I typed, but yeah, the answer should be positive 8/21.