Results 1 to 6 of 6

Math Help - Hard Series Question

  1. #1
    Junior Member
    Joined
    Apr 2010
    Posts
    29

    Exclamation Hard Series Question

    Hi

    For this question:

    The sequences {1, 3, 5, ,
    p} and {1, 3, 5, , q} contain the integer values of

    p
    and q respectively.

    Find the value of
    p+q if :
    {1 + 3 + 5 + + p} +{1 + 3 + 5 + + q} = {1 + 3 + 5 + + 33}

    I've gotten this far:

    0.5 (p+1) [(p+1)/2] + 0.5 (q+1) [(q+1)/2] = 0.5 (33+1) [(33+1)/2]

    ...

    (p+1)^2 + (q+1)^2 = 34^2

    What do I do now to get p + q???

    Thank you!


    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,492
    Thanks
    1393
    You are correct that (p + 1)^2 + (q + 1)^2 = 34^2.

    However, there is not any Pythagorean Triad which satisfies this equation. So I would say that this is unsolvable.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,492
    Thanks
    1393
    Quote Originally Posted by kashuv View Post
    hi... i just joine this site... could u guide me as to how to start a thread? would be very greatful...thanks
    In a sub-forum, try pressing the "Post New Thread" link.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jul 2009
    Posts
    192
    Thanks
    4
    16^2+30^2=34^2

    so p=15 q=29
    Last edited by Krahl; July 25th 2010 at 04:58 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by UltraGirl View Post
    Hi

    For this question:

    The sequences {1, 3, 5, ……,
    p} and {1, 3, 5, ………, q} contain the integer values of

    p
    and q respectively.

    Find the value of
    p+q if :
    {1 + 3 + 5 + ……+ p} +{1 + 3 + 5 + …… + q} = {1 + 3 + 5 + ……+ 33}

    I've gotten this far:

    0.5 (p+1) [(p+1)/2] + 0.5 (q+1) [(q+1)/2] = 0.5 (33+1) [(33+1)/2]

    ...

    (p+1)^2 + (q+1)^2 = 34^2

    What do I do now to get p + q???

    Thank you!


    You could try...

    sum of an arithmetic series is......... average(number\ of\ terms)

    1+3+5+....+p=\left(\frac{p+1}{2}\right)^2

    1+3+5+....+q=\left(\frac{q+1}{2}\right)^2

    1+3+5+....+33=17+17+17+...+17=17^2

    Now we can ask if 17^2 is the sum of 2 squares.

    You can do this by drawing a 17 by 17 square.
    Then if we isolate a 16 by 16 square inside it, we are left with 1(17)+1(16)=33

    If we isolate a 15 by 15 square, we have 2(17)+2(15)=34+30=64

    That's a square, so we have 17^2=15^2+8^2

    Hence p (or q) is 29... from \frac{p+1}{2}=15

    and the other (in this case q) is \frac{q+1}{2}=8\ \Rightarrow\ q=15

    Hence p+q=44 is a solution, for p=29 and q=15.
    However, that will not hold for other values of p+q that give 44 obviously,
    since clearly p or q cannot be >33 among other things.


    you can try a few more combinations to see if there are other solutions

    17^2=14^2+3(17)+3(14)

    17^2=13^2+4(17)+4(13)

    17^2=12^2+5(17)+5(12) etc



    crikey!! Krahl got there first, was i really that long typing this??
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,707
    Thanks
    627
    Hello, UltraGirl!


    Find the value of p+q if

    (1\!+\!3\!+\!5\!+ \hdots +\!p) + (1\!+\!3\!+\!5\!+ \hdots +\!q) \:=\: (1\!+\!3\!+\!5\!+ \hdots +\!33)

    Fact: .The sum of the first n odd numbers is n^2.


    We have:

    . . \underbrace{(1\!+\!3\!+\!5\!+ \hdots +\!p)}_{\frac{p+1}{2}\text{ terms}}  + \underbrace{(1\!+\!3\!+\!5\!+ \hdots +\!q)}_{\frac{q+1}{2}\text{ terms}} \:=\: \underbrace{(1\!+\!3\!+\!5\!+ \hdots +\!33)}_{\text{17 terms}}


    Hence: . \left(\dfrac{p+1}{2}\right)^2 + \left(\dfrac{q+1}{2}\right)^2 \;=\;17^2


    We have a Pythagorean Triple: . a^2 + b^2 \:=\:17^2

    . . and the only solution is: . \{a,b\} \:=\:\{8,15\}


    So we have: . \begin{Bmatrix}\dfrac{p+1}{2} &=& 8 && \Rightarrow && p &=& 15 \\ \\[-3mm] \dfrac{q+1}{2} &=& 15 && \Rightarrow && q &=&29 \end{Bmatrix}


    Therefore: . p + q \:=\:15 + 29 \:=\:44

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. hard question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 25th 2010, 04:31 PM
  2. Hard question.
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: December 8th 2009, 12:48 PM
  3. Hard Fourier Series : Exponentials
    Posted in the Calculus Forum
    Replies: 16
    Last Post: September 15th 2009, 11:32 PM
  4. hard calculas series :(
    Posted in the Calculus Forum
    Replies: 6
    Last Post: April 24th 2008, 08:51 AM
  5. Need some help with a hard question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 23rd 2007, 05:05 AM

Search Tags


/mathhelpforum @mathhelpforum