How does this algebraically simplify to reach this result? A problem I saw in my textbook uses the following logic to simplify an equation.
$$\frac{m}{2}(m+1)+(m+1)\cdot\frac{2}{2}$$
$$=\frac{m+1}{2}(m+2)$$
$$=\frac{m+1}{2}[(m+1)+1]$$
The problem that I'm having is that I do not understand how the first line simplifies to the second line. Whenever I attempt to simplify the problem I am left with $$\frac{m(m+1)}{2}+m+1$$ which is pretty much the same as if the 2/2 was never included in the problem. Any insight on how this works would be greatly appreciated.
 A: $$\frac{m}{2}(m+1)+(m+1)\cdot\frac{2}{2}= \\\frac{m+1}2\cdot m+\frac{m+1}2\cdot2= \\\frac{m+1}2\cdot (m+2)$$
where in the first we rearrange the terms on each of the plus sign and in the second we use the distributive law.
A: In addition to the previous answers, it helps to consider the ( m + 1 ) as a single unit, u.
$$\frac{m}{2}(m+1)+(m+1)\cdot\frac{2}{2}$$
becomes
$$\frac{m}{2}\cdot  u + u \cdot\frac{2}{2}=$$
$$u\cdot(\frac{m}{2} + \cdot\frac{2}{2})=$$
and finally
$$u\cdot(\frac{m+2}{2})$$
Related note: Factoring by grouping frequently uses this concept of factoring a non-trivial variable or expression .
A: The $\frac{2}{2}$ was included as a little hint about the symmetry of the two terms.  They both have something over 2 times $(m+1)$.  This suggests that we factor out the $\frac{(m+1)}{2}$.  When we do that we are left with
$$\frac{m}{2}(m+1)+\frac{2}{2}(m+1)$$
$$m(\frac{m+1}{2}) + 2(\frac{m+1}{2})$$
$$(\frac{m+1}{2})(m+2)$$
A: According to symmetry , Take $\frac{m+1}{2}$ as a factor . $$\frac{m}{2}(m+1)+\frac{2}{2}(m+1)$$ $$\left(\frac{(m+1)(m+2)}{2}\right)$$
