Originally Posted by
math61688 (k/k+1)+ (1/(k+1)(k+2)) = (k+1)/(k+1)+1
The right side must be $\displaystyle \frac{k+1}{(k+1)+1}=\frac{k+1}{k+2}$ , but then:
$\displaystyle \frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{1}{k+1}\l eft(k+\frac{1}{k+2}\right)=\frac{1}{k+1}\left(\fra c{k^2+2k+1}{k+2}\right)$ $\displaystyle =\frac{1}{k+1}\cdot \frac{(k+1)^2}{k+2}=\frac{k+1}{k+2}$
Tonio
I just need to make the left side equal to the right side. This is what I have:
k(k+2)/(k+1)(k+2) + 1/(k+1))k+2) =(k+1)/(k+1)+1
k(k+2)+1/(k+1)(k+2) = (k+1)/(k+1)+1