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    • Thread Starter


    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

    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....)
    • Thread Starter

    Hi JamesF!

    I still don't understand the second part!

    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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