Results 1 to 3 of 3

Math Help - Really hard pre cal problems (4)~(5)

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    13

    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)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,713
    Thanks
    633
    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 \\<br />
a_2 &=& i(2i) &=& \text{-}2 \\<br />
a_3 &=& \text{-}2(2i) &=& \text{-}4i \\<br />
a_4 &=& \text{-}4i(2i) &=& 8 \\<br />
a_5 &=& 8(21) &=& 16i \\ \hline \\[-4mm]<br />
& &\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}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    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)
    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

    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 3 Bad Hard Problems
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 21st 2009, 01:46 PM
  2. 4 hard problems please help
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: December 9th 2008, 10:43 PM
  3. Hard Problems
    Posted in the Geometry Forum
    Replies: 1
    Last Post: August 22nd 2008, 10:38 PM
  4. 2 hard word problems
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: June 30th 2008, 10:52 AM
  5. Two Hard Derivitive Problems
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 25th 2007, 07:02 PM

Search Tags


/mathhelpforum @mathhelpforum