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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

$$\frac{2k }{ k + 1} + \frac{1}{1+2+3+...+(k+1)}= \frac{2(k + 1) }{ (k + 1) + 1}$$

Alaa Mhmd
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  • both sides of what? – Chaos Mar 01 '21 at 18:38
  • I posted a picture but it's not appearing – Alaa Mhmd Mar 01 '21 at 18:40
  • Are you trying to solve an equation or prove an identity? – Joe Mar 01 '21 at 18:44
  • Prove an identity – Alaa Mhmd Mar 01 '21 at 18:45
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    What you've written is not an identity. It might help if you go back to the beginning of the problem you're trying to solve and state the original question. (The image attached to your initial post shows two steps leading to the equation you're asking about here. Those steps both look correct. But where did the top line in the image come from?) – Barry Cipra Mar 01 '21 at 18:50
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    Please check the identity. As written, it's clearly false: the l.h.s. has $0$ in the denominator for $k=0$, whereas the r.h.s. does not. – Bernard Mar 01 '21 at 18:51

3 Answers3

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$1+2+\cdot+(k+1)={(k+1)(k+2)\over2}$, 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.

Barry Cipra
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  • Thank you I solved it now with your help. Thanks a lot. – Alaa Mhmd Mar 01 '21 at 19:09
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    Excellent, I'm glad to have helped. But if you plan to make regular use of this site (and there is no reason not to!), please note how it helps to give a complete description of the problem you're trying to solve and the steps you taken yourself. It might not hurt, in fact, as an exercise here to edit your question to start with the problem and then show the steps that were in your image, so that readers who come to the page not having seen it before can see what you did (and why my answer is pertinent). – Barry Cipra Mar 01 '21 at 19:18
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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.

Simone
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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.

user247327
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