Posted by Aliyah on Tuesday, August 23, 2011 at 1:06am.
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!
Thank you so much!!!
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