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Mathematical Induction

  1. Aug 18, 2005 #1
    Question

    Suppose [itex]a,d \in \mathbb{Z}[/itex] and consider the arithmetic sequence [itex]\{a+kd\}_{k\in\mathbb{N}\cup\{0\}}[/itex]. Use the Principle of Mathematical Induction to prove that

    [tex]\sum_{k=0}^n a+kd = \frac{1}{2}(n+1)(2a+nd)[/tex]
     
    Last edited: Aug 18, 2005
  2. jcsd
  3. Aug 18, 2005 #2
    [1] try it for the base case(guess its n=1 not n=0 summation starts at 1)
    [2] assume it for some n-1 OR n
    [3] show that [2] leads to n OR n+1
    (depends on if your more comfortable n-1 --> n OR n-->n+1
    majority of cases they are the same but sometimes its easier to see one over the other)

    I suggest if you want more help that you first show us what you've done first.
     
  4. Aug 18, 2005 #3
    You're too quick Neurocomp :smile: , I was typing my solution.

    Solution

    Let [itex]P(n)[/itex] be that statement that

    [tex]\sum_{k=0}^n a+kd = \frac{1}{2}(n+1)(2a+nd) \quad \dag[/tex]

    1. Show that [itex]P(1)[/itex] is true:

    [tex]P(1) = \sum_{k=0}^1 a+kd = (a+0d)+(a+1d) = 2a + d \quad \ddag[/tex]

    For [itex]n=1[/itex] the RHS of [itex]\dag[/itex] equals

    [tex]\frac{1}{2}(1+1)(2a + 1d) = 2a+d = \ddag[/tex]

    2. Assume true for [itex]P(n)[/itex].

    3. Prove true for [itex]P(n+1)[/itex]. That is [itex]P(n+1) = \frac{1}{2}(n+2)(2a+(n+1)d)[/itex]:

    [tex]\sum_{k=0}^{n+1} a+kd &=& \sum_{k=0}^n a+kd + \sum_{k=n+1}^{n+1} a+kd[/tex]
    [tex]\quad= \sum_{k=0}^n a+kd + [a + d(n+1)][/tex]
    [tex]\quad= \frac{1}{2}(n+1)(2a+nd) + a + d(n+1)[/tex]
    [tex]\quad= a(n+1) + a + \frac{1}{2}dn(n+1) + d(n+1)[/tex]
    [tex]\quad= a(n+2) + \frac{1}{2}d(n+1)(n+2)[/tex]
    [tex]\quad= \frac{1}{2}(n+2)(2a + (n+1)d)[/tex]
    [tex]\quad= P(n+1)[/tex]

    [tex]\square[/tex]
     
    Last edited: Aug 18, 2005
  5. Aug 18, 2005 #4
    It may be hard to see what I've done in going from

    [tex]a(n+2) + a + \frac{1}{2}dn(n+1) + d(n+1)[/tex]

    to the next line.

    All Ive done is group the a and [itex]\frac{1}{2}d(n+1)[/tex]

    So

    [tex]\frac{1}{2}dn(n+1) + d(n+1) = \frac{1}{2}\left[dn(n+1) + 2d(n+1)\right][/tex]
    [tex] = \frac{1}{2}d\left[n(n+1) + 2(n+1)\right][/tex]
    [tex] = \frac{1}{2}d(n+1)\left[n+2\right][/tex]
     
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