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Posted by on Tuesday, August 23, 2011 at 1:06am.

Prove by mathematical induction that
: E (3r-5)= 3n^2-7n /2
r=1

  • math - , Tuesday, August 23, 2011 at 8:14am

    is that ∑(3r-5) as r goes from 1 to n ??

    I will asssume it is

    then ∑(3r-5) = (3n^2 - 7n)/2

    Step 1 : test for n=1
    LS = 3(1) - 5 = -2
    RS = (3(1) - 7)/2 = -2

    Step 2 : assume it true for n = k, that is ..

    -2 + 1 + 4 + ... + (3k-5) = (3k^2 - 7k)/2

    Step 3: prove it to be true for n = k+1
    that is, prove
    -2 + 1 + 4 + ... + (3k-5) + (3(k+1)-5) = (3(k+1)^2 - 7(k+1))/2

    LS = (3k^2 - 7k)/2 + 3(k+1) - 5
    = (3k^2 - 7k)/2 + 3k + 3 - 5
    = (3k^2 - 7k + 6k - 4)/2
    = (3k^2 - k - 2)/2

    RS = (3(k+1)^2 - 7(k+1))/2
    = (3k^2 + 6k + 3 - 7k - 7)/2
    = (3k^2 - k -4)/2

    Yeahhh!

  • math - , Tuesday, August 23, 2011 at 5:30pm

    Thank you so much!!!

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