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)]
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?
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)}\,.
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 - ...