Recent content by gcarson1

A yes! Brilliant, the A made things so much clearer, I resolved the other problem in class today while reading this, thankyou very much!

line 1 1x1!+…+nxn!+((n+1)(n+1)!)=[ 1x1!+…+nxn! ] + ((n+1)(n+1)!) line 2 1x1!+…+nxn!+((n+1)(n+1)!)= [(n+1)!-1]+((n+1)(n+1)!) line 3 1x1!+…+nxn!+((n+1)(n+1)!)= [(n+1)!-1]+((n+1)(n+1)!) So the next line should be line 4 1x1!+…+nxn!+((n+1)(n+1)!)=(n+2)!-1 I see logically why that makes sense...

Ok I think I got it, the expansion is a little messy but here goes : Prove that 3|n^3+2n whenever n is a positive integer. Basis Step: 1^3+2(1)/3 = 3/3 = 1 then we know that the first step in the formula is true. Inductive Step: 3|(n+1)^3+2(n+1) If both terms are multiples of 3 then they are...

Tried to divide both by 3 in order to redue the formula to a positive integer. Yet I still don't understand how to write or show this?

Yes thankyou, I was copying and pasting and missed that line. Thankyou still for pointing that out. I'm still confused that, as all I have done than is repeat the line in the geometric progression on the right side of the equation. Ex. 1x1!+…+nxn!+((n+1)(n+1)!)= ((n+1)(n+1)!) the right side...

Hi all, been struggling with my course material regarding Induction. My prof is really not great at explaining this proof method. I don't know what the learning curve or expectations here are like but I'm really struggling with this conceptually. Any assistance with these problems would be much...

Hi all, been struggling with my course material regarding Induction. My prof is really not great at explaining this proof method. I don't know what the learning curve or expectations here are like but I'm really struggling with this conceptually. Any assistance with these problems would be much...