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3 math questions (summation, mathematical induction)

  1. Apr 24, 2005 #1
    2 math questions (summation, mathematical induction)

    I have 2 questions regarding summation and mathematical induction

    2. Prove by mathematical induction
    [tex] \sum^n_{r=1} \frac {1}{r(r+2)} = \frac {3}{4} - \frac {(2n+3)}{2(n+1)(n+2)} [/tex]
    i am now trying to prove that

    [tex] 3/4 - \frac{2(n+1)+3}{2(n+2)(n+3)} - \frac {1}{n+1}{n+3} [/tex]

    I seem to get caught up with the algebra and what i get is

    [tex] \frac {2n^2+5n+1}{2(n+1)(n+2)(n+3)} [/tex]. instead of [tex] \frac {2n^2+5n+3}{2(n+1)(n+2)(n+3)} [/tex]

    3. Prove by induction that [tex] \sum^n_{r=1}rx^{r-1}=\frac{1-(n+1)x^n+nx^{n+1}}{(1-x)^2} [/tex]

    again, i'm caught up with the algebra.
     
    Last edited: Apr 24, 2005
  2. jcsd
  3. Apr 24, 2005 #2
    I use a different method:

    1) proof summation is true when n=2
    2) consider n=k+1.
    [tex]\sum^{k+1}_{r=1} \frac{1}{r(r+2)} = \frac{3}{4} - \frac{2(k+1)+3}{2(k+2)(k+3)}[/tex]
    [tex]= \sum^k_{r=1} \frac{1}{r(r+2)} + \mbox{(k+1)^{th}} \mbox{term}[/tex]

    Well, it works. Sorry if I'm not very helpful :uhh:
     
  4. Apr 24, 2005 #3

    HallsofIvy

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    "i am now trying to prove that
    [tex] 3/4 - \frac{2(n+1)+3}{2(n+2)(n+3)} - \frac {1}{n+1}{n+3} [/tex]"

    This makes no sense- it is neither an equation nor a statement- it is a formula. What are you trying to prove about it?

    What you should be trying to prove is that
    [tex]\frac {3}{4} - \frac {(2n+3)}{2(n+1)(n+2)}+ \frac{1}{(n+1)(n+3)} = \frac{3}{4}- \frac{(2n+5)}{2(n+2)(n+3)}[/tex]
    The left side of that is just the formula for the sum to n plus the n+1 term and the right side is the same formula for the sum to n+1.
    Now to show they are the same: obviously the "3/4" cancel. Get the common denominator for the two fractions left on the left side:
    [tex]-\frac {(2n+3)(n+3)}{2(n+1)(n+2)(n+3)}+ \frac{2(n+2)}{2(n+1)(n+2)(n+3)}[/tex]
    [tex]-\frac{(2n^2+ 9n+ 9)+(2n+4)}{2(n+1)(n+2)(n+3)}[/tex]
    [tex]-\frac{2n^2+7n+5}{2(n+1)(n+2)(n+3)}[/tex]
    In order to have the same denominator as on the right, we obviously need to cancel the "n+1" in the denominator and, sure enough, 2n2+ 7n+ 5= (n+1)(2n+ 5)!

    As far as 2 is concerned, are you required to use induction? I notice that the left side is just the derivative of 1+ x+ x2+...+ xn and I know that xn+1- 1= (x- 1)(1+ x+ x2+...+ xn) so that
    [tex]1+ x+ x^2+...+ x^n= \frac{x^n-1}{x-1}[/tex]. You can get the result by using the quotient rule to differentiate the right side.

    Of course, if you MUST us induction, then you need to show that
    [tex]\frac{1-(n+1)x^n+nx^{n+1}}{(1-x)^2}+ (n+1)x^n= \frac{1-(n+2)x^2+(n+1)x^{n+2}}{(1-x)^2} [/tex]

    Again, concetrate on the right hand side. You will need to multiply that "(n+1)xn" term by (1-x)2/(1-x)2 and then combine the numerators.
     
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