# Thread: Really hard pre cal problems (4)~(5)

1. ## Really hard pre cal problems (4)~(5)

(4) Terms of a sequence can also be complex numbers. All the usual rules and formulae apply.
a. If the first term of a geometic sequence is i and the common ratio is r=2i, list the first five terms of this sequence and find their sum.
b. Identify r and find the sum of this infinite geometric series:
27+(-9i)+(3i^2)+(-i^3)+((i^4)/3).....

(5)
a. There is a sequence u with an associated series S, and all we know is that S(n)=n^3. Is the sequence u arithmetic, geometric, or neither?
b. Find a formula for u(n), the nth term of the sequence. Hint: S(n)=S(n-1)+u(n)

2. Hello,vvc531!

(4) Terms of a sequence can also be complex numbers.
All the usual rules and formulae apply.

a. If the first term of a geometic sequence is $i$ and the common ratio is $r=2i$,
list the first five terms of this sequence and find their sum.

$\begin{array}{ccccc}a_1 &=& i &=& i \\
a_2 &=& i(2i) &=& \text{-}2 \\
a_3 &=& \text{-}2(2i) &=& \text{-}4i \\
a_4 &=& \text{-}4i(2i) &=& 8 \\
a_5 &=& 8(21) &=& 16i \\ \hline \\[-4mm]
& &\text{Total:} & & 6 + 13i \end{array}$

b. Identify $r$ and find the sum of this infinite geometric series:
. . . $S \;=\;27 -9i + 3i^2 -i^3 + \frac{i^4}{3} + \hdots$

We have: . $a_1 = 27,\;a_2 = -9i$

Since $r \:=\:\frac{a_2}{a_1}$, we have: . $r \:=\:\frac{\text{-}9i}{27} \:=\:-\frac{1}{3}i$

Since $S \:=\:\frac{a_1}{1-r}$ .we have: . $S \;=\;\frac{27}{1 - \left(\text{-}\frac{1}{3}i\right)} \;=\;\frac{27}{1 + \frac{1}{3}i} \;=\;\frac{81}{3 + i}$

Rationalize: . $\frac{81}{3+i}\cdot\frac{3-i}{3-i} \;=\;\frac{81(3-i)}{10}$

3. ## Series

Hello vvc531
Originally Posted by vvc531
(5)
a. There is a sequence u with an associated series S, and all we know is that S(n)=n^3. Is the sequence u arithmetic, geometric, or neither?
b. Find a formula for u(n), the nth term of the sequence. Hint: S(n)=S(n-1)+u(n)
$S(1) = 1^3 = 1 = u(1)$

$S(2)= 2^3 = 8 \Rightarrow u(2) = 7$

$S(3) = 3^3 = 27 \Rightarrow u(3) = 19$

Differences of consecutive terms are $7-1=6$ and $19-7=12$. So it's not arithemetic.

Ratios of consecutive terms are $\frac{7}{1}=7$ and $\frac{19}{7}$. So it's not geometric either.

$S(n) = n^3, S(n-1) = (n-1)^3$ and $S(n) = S(n-1) + u(n)$

$\Rightarrow u(n) = n^3 - (n-1)^3$

$= 3n^2 -3n+1$