Find the formula of
n
Σ (j+3)(j-4)
j=1
I have gotten to
n(n+1)(2n+1)/6 - n(n+1)/2 - 12n
How do I simplify it from there?
see your other post. Repeating it does no one any good.
Don't forget your Algebra I now that you're taking Calculus!
n
Σ (j+3)(j-4)
j=1
I have gotten to
n(n+1)(2n+1)/6 - n(n+1)/2 - 12n
How do I simplify it from there?
Don't forget your Algebra I now that you're taking Calculus!