Question Consider the following sequence,
$3,1,{\frac {1}{3}},{\frac {1}{9}},{\frac {1}{27}},\cdots$
a. Determine a formula for $a_{n}$, the nth term of the sequence.
b. Find the sum $\displaystyle {\sum _{k=1}^{\infty }a_{k}}$
Foundations

1) What type of series is this?

2) Which formulas, about this type of series, are relevant to this question?

3) In the formula there are some placeholder variables. What is the value of each placeholder?

Answer:

1) This series is geometric. The giveaway is there is a number raised to the nth power.

2) The desired formulas are $a_{n}=a\cdot r^{n1}$ and $S_{\infty }={\frac {a_{1}}{1r}}$

3) $a_{1}$ is the first term in the series, which is $3$. The value for r is the ratio between consecutive terms, which is ${\frac {1}{3}}$

Step 1:

The sequence is a geometric sequence. The common ratio is $r={\frac {1}{3}}$.

Step 2:

The formula for the nth term of a geometric series is $a_{n}=ar^{n1}$ where $a$ is the first term of the sequence.

So, the formula for this geometric series is $a_{n}=(3)\left({\frac {1}{3}}\right)^{n1}$.

Step 3:

For geometric series, $\displaystyle {\sum _{k=1}^{\infty }a_{k}}={\frac {a}{1r}}$ if $r<1$. Since $r={\frac {1}{3}}$,

we have $\displaystyle {\sum _{k=1}^{\infty }a_{k}}={\frac {3}{1{\frac {1}{3}}}}={\frac {9}{4}}$.

Final Answer:

$a_{n}=(3)\left({\frac {1}{3}}\right)^{n1}$

${\frac {9}{4}}$

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