How can I prove that the l.h.s equals the r.h.s? I can't move forward.  Can anyone help?
$$\frac{k (2k) + 2 }{ k (k + 1) }= \frac{2k + 2 }{ k + 2}$$
I'm trying to prove that the left side equals the right side.
It started like this
$$\frac{2k }{ k + 1} + \frac{1}{1+2+3+...+(k+1)}= \frac{2(k + 1) }{ (k + 1) + 1}$$
 A: $1+2+\cdot+(k+1)={(k+1)(k+2)\over2}$, not $k(k+1)\over2$. If you make that change in the steps you carried out (in the image attached to the original version of the posted question), you should see that you get an identity.
A: Perhaps you made a typo, or it might be a trick question: for $k=1$ we get $\frac {1}{2} = \frac 43$, which is false.
A: Why would you WANT to "make both sides equal".  That's a very unusual thing to do.
More often people want to "solve for k" or "prove an identity" as Joe suggested.
I would start by trying to "solve for k", If it turns out that k can be any number, then it is an identity.
You have $\frac{k(2k+ 1)}{k(k+1)}= \frac{2k+ 2}{k+2}$.
The first thing you can do is cancel the "k" in both numerator and denominator in the left. That leaves $\frac{2k+ 1}{k+ 1}= \frac{2k+ 2}{k+ 2}$.
Eliminate the fractions (I really don't like fractions!) by multiplying both sides by $(k+ 1)(k+ 2)$
$(2k+ 1)(k+ 2)= (2k+ 2)(k+ 1)$
$2k^2+ 5k+ 2= 2k^2+ 4k+ 2$
Subtract $2k^2+ 4k+ 2$ from both sides to get
k^2= 0.
NO, this is not an identity.  It is only true for k= 0.
