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Math Help - induction problem

  1. #1
    baz
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    induction problem

    Can anybody help me with this.

    Prove that by mathematical induction for n>=1
    1/1.4 + 1/4.7 + ................+ 1/(3n-2).(3n+1) = n/(3n+1)

    thanks in advance.
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  2. #2
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    Quote Originally Posted by baz View Post
    Can anybody help me with this.

    Prove that by mathematical induction for n>=1
    1/1.4 + 1/4.7 + ................+ 1/(3n-2).(3n+1) = n/(3n+1)

    thanks in advance.
    Perhaps the following could help you:
    RHS can be rewritten as \frac13\left(1-\frac1{3n+1}\right).
    \frac1{(3n-2)(3n+1)}=\frac13\left(\frac1{3n-2}-\frac1{3n+1}\right)
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  3. #3
    baz
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    Quote Originally Posted by kompik View Post
    Perhaps the following could help you:
    RHS can be rewritten as \frac13\left(1-\frac1{3n+1}\right).
    \frac1{(3n-2)(3n+1)}=\frac13\left(\frac1{3n-2}-\frac1{3n+1}\right)
    I didnot det you.How come R.H.S be rewritten as you said?
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  4. #4
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    Quote Originally Posted by baz View Post
    I didnot det you.How come R.H.S be rewritten as you said?
    Just try to use the common denominator:
    \frac13\left(1-\frac1{3n+1}\right)=<br />
\frac13\left(\frac{3n+1}{3n+1}-\frac1{3n+1}\right)=<br />
\frac13\cdot\frac{3n}{3n+1}=\frac{n}{3n+1}<br />
    which is exactly the RHS.
    But the modification of the RHS is perhaps not so important.

    Just try to use:
    \frac1{(3n-2)(3n+1)}=\frac13\left(\frac1{3n-2}-\frac1{3n+1}\right)
    or (the same equation with n+1 instead of n
    \frac1{(3n+1)(3n+4)}=\frac13\left(\frac1{3n+1}-\frac1{3n+4}\right)
    in the inductive step of your proof.

    EDIT: BTW the sums of this type are called telescopic. With a little practice you can recognize easily at least the simple ones - and then you see the result immediately (that was the reason why I rewrote the RHS - since I saw that this expression is what I get from the telescopic sum.
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  5. #5
    baz
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    Quote Originally Posted by kompik View Post
    Just try to use the common denominator:
    \frac13\left(1-\frac1{3n+1}\right)=<br />
\frac13\left(\frac{3n+1}{3n+1}-\frac1{3n+1}\right)=<br />
\frac13\cdot\frac{3n}{3n+1}=\frac{n}{3n+1}<br />
    which is exactly the RHS.
    But the modification of the RHS is perhaps not so important.

    Just try to use:
    \frac1{(3n-2)(3n+1)}=\frac13\left(\frac1{3n-2}-\frac1{3n+1}\right)
    or (the same equation with n+1 instead of n
    \frac1{(3n+1)(3n+4)}=\frac13\left(\frac1{3n+1}-\frac1{3n+4}\right)
    in the inductive step of your proof.

    EDIT: BTW the sums of this type are called telescopic. With a little practice you can recognize easily at least the simple ones - and then you see the result immediately (that was the reason why I rewrote the RHS - since I saw that this expression is what I get from the telescopic sum.
    thanks man,this is helpful.
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