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    Hi

    This is a two part question and part A is easy but part B I don't understand!

    a) Prove by induction that, for n E N, the sum of the first n integers is given by the formula n(n+1)/2

    ANSWER I can find!

    b) Find the value of (1+2) + (4+5) + (7+8) +......+ [(3n-2)+(3n-1)] in terms of n

    ANSWER I don't know how to do this part! I do know however that part A has something to do with the solution to this!

    Please explain! Thanks
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    Show that its true for n=1 and then assume its true for n.
    To prove it is true for n+1, we take the sum from 1 to n, and then add the next number to the end, so we get

    Sum from 1 to (n+1) = n(n+1)/2 + (n+1) = ((n^2 + n) + 2n + 2)/2 = (n^2 + 3n + 2)/2 = (n+1)(n+2)/2, which proves it.

    For the second part, notice that the sum is all the numbers from 1 to n, except those divisible by 3. And that the sum of the numbers taken out is
    3 + 6 + 9 + 12.... = 3 (1 + 2 + 3 + 4....)
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    Hi JamesF!

    I still don't understand the second part!
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    The second part is the same as the sum of all the numbers up to 3n and then taking away all the multiples of 3.

    So you get
    (3n)(3n+1)/2 - (3+6+9+12.....+3n)

    Take out a factor of 3 in the second part and you get
    (3n)(3n+1)/2 - 3(1+2+3+4.....+n)

    (3n)(3n+1)/2 - 3n(n+1)/2
 
 
 
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Updated: July 21, 2004

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