Recent content by Lococard

(n+2) / (n+2) - [1/(n+2)]? = [(n+2) - 1] / (n+2)?

Im really not sure. You can cancel the (n+1) / (n+1) Leaving (n+1) / (n+2)? Im a little confused as the teacher suggested that i work my way to get 1 - [1 / (n+2)]

So that is the same as: \frac{(n+1)(n+1)}{(n+1)(n+2)}? yes? no? Now I am not sure what to do :S

haha, Sorry i lost my password and this connection doesn't allow gmail.com to load. But I am back now. \frac{n(n+2)+1}{(n+1)(n+2)} And from then i simplfy it?

\frac{n}{(n+1)} + \frac{1}{(n+1)(n+2)} = 0 n/(n+1) = n(n+2)/(n+1)(n+2).Did you swap \frac{1}{(n+1)(n+2)} to the other side of the equals sign? How did you simplify did you get n(n+2)/(n+1)(n+2). Is it all on the LHS of the equals sign, or did you swap some of it to the right side?

how do you transfer \frac{1}{(n+1)(n+2)} from the LHS to the RHS?

That was the bit i was stuck on.

Ah. 1 - \frac{1}{N+1} = \frac{n+1}{n+1} - \frac{1}{n+1} = \frac{n}{n+1}How did you get: \frac{n+1}{(n+1)} - \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)} to get to: \frac{n+1}{(n+1)} - \frac{1}{(n+1)}\,=\,\frac{n}{(n+1)}\,.

How did you do that step? Now I am really lost.The lecturer suggested i get the equation to = 1 - \frac{1}{n+2}

did you just substitute the 1 for a \frac{n+1}{n+1}?Im really stuck on simplifying the RHS.

How would i do the next part?

Oh whoops. Yeah that's what i meant

Alright, i got some more info on the question. It seems that: \frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}... + \frac{1}{n+1} = 1-\frac{1}{n+1} But. The number after \frac{1}{n+1} would be \frac{(n+1)}{(n+1)(n+2)} So. 1 - \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)} = 1 - ...

I've done them both on paper and i got the same result. Do i then add a_n+1 to the equation and get the formula in the lowest form?

How would i put it into a partial fraction?