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September 30, 2014

September 30, 2014

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

: E (3r-5)= 3n^2-7n /2

r=1

- math -
**Reiny**, Tuesday, August 23, 2011 at 8:14amis 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 -
**Aliyah**, Tuesday, August 23, 2011 at 5:30pmThank you so much!!!

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