Posted by Aliyah on Tuesday, August 23, 2011 at 1:06am.
Prove by mathematical induction that
: E (3r5)= 3n^27n /2
r=1

math  Reiny, Tuesday, August 23, 2011 at 8:14am
is that ∑(3r5) as r goes from 1 to n ??
I will asssume it is
then ∑(3r5) = (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 + ... + (3k5) = (3k^2  7k)/2
Step 3: prove it to be true for n = k+1
that is, prove
2 + 1 + 4 + ... + (3k5) + (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  Aliyah, Tuesday, August 23, 2011 at 5:30pm
Thank you so much!!!