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

  1. #1
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    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)
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  2. #2
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    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 $\displaystyle i$ and the common ratio is $\displaystyle r=2i$,
    list the first five terms of this sequence and find their sum.

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



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

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

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


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


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

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  3. #3
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    Series

    Hello vvc531
    Quote Originally Posted by vvc531 View Post
    (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)
    $\displaystyle S(1) = 1^3 = 1 = u(1)$

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

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

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

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

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

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

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

    Grandad
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