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Math Help - differences or induction

  1. #1
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    differences or induction

    Prove, using the methods of differences or induction, that

    1 / b + a / b(b+1) + ... + a(a+1) ... (a + n - 1) / b(b+1)...(b+n) =

    1 / (b - a) { 1 - a(a+1)...(a+n) / b(b+1)...(b+n)}

    All the best,

    Joe
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Joe5000 View Post
    Prove, using the methods of differences or induction, that

    1 / b + a / b(b+1) + ... + a(a+1) ... (a + n - 1) / b(b+1)...(b+n) =

    1 / (b - a) { 1 - a(a+1)...(a+n) / b(b+1)...(b+n)}

    All the best,

    Joe
    For n = 1 this reads:
    \frac{1}{b} + \frac{a}{b(b + 1)} = \frac{1}{b - a} \frac{1 - a(a + 1)}{b(b + 1)}
    which is not true.

    -Dan
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  3. #3
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    differences or induction

    For n = 1 this reads,
    LHS 1 / b + a / b(b+1) = b+1 + a / b(b+1)
    RHS 1 / b-a { 1 - a(a+1) / b(b+1) } no need for extra brackets
    = 1 / b-a [ b(b+1) - a(a+1 / b(b+1) ] notice the change in external brackets now that the expression is being worked
    = 1 / b-a [ b^2 + b - a^2 - a / b(b+1) ]
    = 1 / b-a [ (b+a)(b-a) + (b-a) / b(b+1) ]
    = 1 / b-a [ (b-a) [ b + a + 1] / b(b+1) ]
    = b + a + 1 / b(b+1)
    ergo LHS = RHS when n = 1. I didn't think that I would have to give basic algebra lessons here. May I suggest one reads that original post carefully!

    ps methods should read method -- a slip of the finger!
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  4. #4
    Forum Admin topsquark's Avatar
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    Angry

    Quote Originally Posted by Joe5000 View Post
    For n = 1 this reads,
    LHS 1 / b + a / b(b+1) = b+1 + a / b(b+1)
    RHS 1 / b-a { 1 - a(a+1) / b(b+1) } no need for extra brackets
    = 1 / b-a [ b(b+1) - a(a+1 / b(b+1) ] notice the change in external brackets now that the expression is being worked
    = 1 / b-a [ b^2 + b - a^2 - a / b(b+1) ]
    = 1 / b-a [ (b+a)(b-a) + (b-a) / b(b+1) ]
    = 1 / b-a [ (b-a) [ b + a + 1] / b(b+1) ]
    = b + a + 1 / b(b+1)
    ergo LHS = RHS when n = 1. I didn't think that I would have to give basic algebra lessons here. May I suggest one reads that original post carefully!

    ps methods should read method -- a slip of the finger!
    I find it fascinating that I am being told I don't know basic algebra by someone who doesn't know how to use parenthesis correctly.

    However, you are correct about one thing: I did read the RHS incorrectly. Typically on this forum when people post the expression:
    w - x/y - z
    they mean
    \frac{w - x}{y - z}
    which does not correspond to order of operations.

    And before you try throwing any more stones, look at the following lines from your response:
    Quote Originally Posted by Joe5000 View Post
    RHS 1 / b-a { 1 - a(a+1) / b(b+1) } no need for extra brackets
    = 1 / b-a [ b(b+1) - a(a+1 / b(b+1) ] notice the change in external brackets now that the expression is being worked
    = 1 / b-a [ b^2 + b - a^2 - a / b(b+1) ]
    = 1 / b-a [ (b+a)(b-a) + (b-a) / b(b+1) ]
    = 1 / b-a [ (b-a) [ b + a + 1] / b(b+1) ]
    = b + a + 1 / b(b+1)
    in which every line makes the same error I assumed in the first place.

    Anyone can make an error. Look to yourself for that as well.

    You are correct about one more thing: someone else is going to have to help you. I don't like your attitude toward someone who, after all, was only trying to help.

    -Dan
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  5. #5
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    By the conventions my friends and I use (and everyone else in college)
    w - x/y - z is understood as

    (w - x) / (y - z).

    Apologies if my use of brackets is non-standard. It is the presence of the minus sign which forces the brackets taking precedence over / by our use. (Same idea for + , i.e. w + x/y - z would be written out as (w + x) / (y + z)

    I will change my use of brackets from now on when posting, including many more, to clarify the expressions used as much as possible.

    All the best,

    Joe
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Joe5000 View Post
    By the conventions my friends and I use (and everyone else in college)
    w - x/y - z is understood as

    (w - x) / (y - z).

    Apologies if my use of brackets is non-standard. It is the presence of the minus sign which forces the brackets taking precedence over / by our use. (Same idea for + , i.e. w + x/y - z would be written out as (w + x) / (y + z)

    I will change my use of brackets from now on when posting, including many more, to clarify the expressions used as much as possible.

    All the best,

    Joe
    Really? Order of operations (PEDMAS or whatever acronym you prefer) says that w + x/y - z should be w + \frac{x}{y} - z. I know a lot of students we see coming on to the forum seem to use the convention you do, but I've never heard of it as a standard. What school are you at?

    Anyway, now that my mistake in my first reply is evident, let me see what I can do with this:

    For the n = 1 case we have
    \frac{1}{b} + \frac{a}{b(b + 1)} = \frac{1}{b - a} \cdot \left ( 1 - \frac{a(a + 1)}{b(b + 1)} \right )
    is correct. (As you showed in your other post, so I won't repeat it here.)

    Now let's assume the theorem is true for some n = k. That is, presume that
    \frac{1}{b} + \frac{a}{b(b+1)} + ~ ... ~ + \frac{a(a+1) ... (a + k - 1)}{b(b+1)...(b+k)} = \frac{1}{b - a} \cdot \left ( 1 - \frac{a(a+1)...(a+k)}{b(b+1)...(b+k)} \right )

    Let's see what the (k + 1)th case has to say. We need to prove that
    \frac{1}{b} + \frac{a}{b(b+1)} + ~ ... ~ + \frac{a(a+1) ... (a + k - 1)}{b(b+1)...(b+k)} + \frac{a(a+1) ... (a + k)}{b(b+1)...(b+(k + 1))} = \frac{1}{b - a} \cdot \left ( 1 - \frac{a(a+1)...(a+(k + 1))}{b(b+1)...(b+(k + 1))} \right )

    From our assumption we have that the first k + 1 terms of the LHS is:
    \frac{1}{b} + \frac{a}{b(b+1)} + ~ ... ~ + \frac{a(a+1) ... (a + k - 1)}{b(b+1)...(b+k)} = \frac{1}{b - a} \cdot \left ( 1 - \frac{a(a+1)...(a+k)}{b(b+1)...(b+k)} \right )

    So inserting the RHS of this expression into the k + 1 equation we get for the LHS:
    \frac{1}{b} + \frac{a}{b(b+1)} + ~ ... ~ + \frac{a(a+1) ... (a + k - 1)}{b(b+1)...(b+k)} + \frac{a(a+1) ... (a + k)}{b(b+1)...(b+(k + 1))}  =  \frac{1}{b - a} \cdot \left ( 1 - \frac{a(a+1)...(a+k)}{b(b+1)...(b+k)} \right ) +  \frac{a(a+1) ... (a + k)}{b(b+1)...(b+(k + 1))}
    (There's no way to get that all on one line. Sorry!)

    So we need to prove that
    \frac{1}{b - a} \cdot \left ( 1 - \frac{a(a+1)...(a+k)}{b(b+1)...(b+k)} \right ) + \frac{a(a+1) ... (a + k)}{b(b+1)...(b+(k + 1))} = <br />
\frac{1}{b - a} \cdot \left ( 1 - \frac{a(a+1)...(a+(k + 1))}{b(b+1)...(b+(k + 1))} \right )

    At this stage I would typically try to add the fractions on the LHS and match the answer to the RHS. That's going to get awfully messy, though. Since all we need to show is that this is an identity let's try this:
    \frac{a(a+1) ... (a + k)}{b(b+1)...(b+(k + 1))}= <br />
\frac{1}{b - a} \cdot \left ( 1 - \frac{a(a+1)...(a+(k + 1))}{b(b+1)...(b+(k + 1))} \right ) - \frac{1}{b - a} \cdot \left ( 1 - \frac{a(a+1)...(a+k)}{b(b+1)...(b+k)} \right )

    \frac{a(a+1) ... (a + k)}{b(b+1)...(b+(k + 1))}= <br />
\frac{1}{b - a} \cdot \left ( 1 - \frac{a(a+1)...(a+(k + 1))}{b(b+1)...(b+(k + 1))} - 1 + \frac{a(a+1)...(a+k)}{b(b+1)...(b+k)} \right )

    \frac{a(a+1) ... (a + k)}{b(b+1)...(b+(k + 1))}= <br />
\frac{1}{b - a} \cdot \left (  \frac{a(a+1)...(a+k)}{b(b+1)...(b+k)} - \frac{a(a+1)...(a+(k + 1))}{b(b+1)...(b+(k + 1))} \right )

    \frac{a(a+1) ... (a + k)}{b(b+1)...(b+(k + 1))}= <br />
\frac{1}{b - a} \cdot \left ( \frac{a(a+1)...(a+k)(b + (k + 1)) - a(a+1)...(a+(k + 1))}{b(b+1)...(b+(k + 1))} \right )

    Now factor the common a(a+1) ... (a + k) in the numerator:
    \frac{a(a+1) ... (a + k)}{b(b+1)...(b+(k + 1))}= <br />
\frac{1}{b - a} \cdot \left ( \frac{((b + (k + 1)) -  (a+(k + 1))) \cdot a(a+1)...(a+k)}{b(b+1)...(b+(k + 1))} \right )

    \frac{a(a+1) ... (a + k)}{b(b+1)...(b+(k + 1))}= <br />
\frac{1}{b - a} \cdot \left ( \frac{(b - a) \cdot a(a+1)...(a+k)}{b(b+1)...(b+(k + 1))} \right )

    \frac{a(a+1) ... (a + k)}{b(b+1)...(b+(k + 1))}= <br />
 \frac{a(a+1)...(a+k)}{b(b+1)...(b+(k + 1))}
    which is an identity, as desired.

    Therefore, since the theorem is true for k = 1, it is true for k = 2, etc.

    -Dan
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  7. #7
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    differences or induction

    Dear Dan,

    Many thanks. You make it all very simple.

    All the best,

    Joe
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