Algebra & Re-arrangement

Question

I need to make this: $K(K+1)(2K+7) + 6(k+1)[(k+1)+2]/6$

Equal to this: $(K+1)[(k+1)+1][(2(k+1)+7]/6$ By using algebra and re-arrangement.

From the initial equation there should be just one more step before the two can equal each-other.

Answer 1

You seem to be asking how to show

$$k(k+1)(2k+7)+{6(k+1)((k+1)+2)\over6}={(k+1)((k+1)+1)(2(k+1)+7)\over6}$$

The problem is, that's not the correct equation. What you really want to show is

$${k(k+1)(2k+7)\over6}+(k+1)((k+1)+2)={(k+1)((k+1)+1)(2(k+1)+7)\over6}$$

Until you are fluent "speaking" algebra (and sometimes even then!), this is best done incrementally, in bite-sized steps:

$$\begin{align} {k(k+1)(2k+7)\over6}+(k+1)((k+1)+2) &={k(k+1)(2k+7)+6(k+1)(k+3)\over6}\\ &={(k+1)\left((2k^2+7k)+(6k+18)\right)\over6}\\ &={(k+1)\left(2k^2+13k+18\right)\over6}\\ &={(k+1)\left(k+2\right)\left(2k+9\right)\over6}\\ &={(k+1)\left((k+1)+1\right)\left(2(k+1)+7\right)\over6}\\ \end{align}$$

One other comment: The mixing of uppercase $K$ and lowercase $k$ in the OP's formulation is inappropriate here, but that's a separate issue which, I think, stems from some uncertainty over how to write an inductive proof, which this algebra question originated in.