Induction proof that $\sum_{j=n}^{2n-1} (2j + 1) = 3n^2$ - what happened? So I have the general summation formula that I was to prove using mathemathical induction on a Calculus level.
For all $n = {1,2,3,...}$ we have:
$$\sum_{j=n}^{2n-1} (2j + 1) = 3n^2$$
In the answer, the one completing the proof gets this, on the $n=k+1$:
$$\sum_{j=(k+1)}^{2(k+1)-1} (2j + 1) = \sum_{j=k}^{2k-1} (2j + 1) - (2k+1) + (4k+1) + (4k+3)$$
$$= 3k^2 + 6k + 3 = 3(k+1)^2$$
Can anyone explain how on earth this was done? When I tried, I was unable to factorize - I tried by adding the $(2(k+1)+1)$ to the $S_n$ formula and thereby confirm it, but I couldn't do it...
 A: Compare the sums
$$\sum_{j=k+1}^{2(k+1)-1}(2j+1)=\sum_{j=k+1}^{2k+1}(2j+1)\tag{1}$$
and
$$\sum_{j=k}^{2k-1}(2j+1)\;,\tag{2}$$
the first being the one that you want to evaluate, the second being the one whose value you know. The first sum does not have a $j=k$ term; the second one does. The first sum does have a $j=2k$ term and a $j=2k+1$ term; the second does not. To convert $(2)$ into $(1)$, therefore, you must subtract the $j=k$ term and add in the extra $j=2k$ and $j=2k+1$ terms. 
The $j=k$ term is $2k+1$; you need to subtract that, because it’s not included in $(1)$. The $j=2k$ and $j=2k+1$ terms are $2(2k)+1=4k+1$ and $2(2k+1)+1=4k+3$; you need to add those, because they’re included in $(1)$ but are not already present in $(2)$. Thus,
$$\sum_{j=k+1}^{2k+1}(2j+1)=\sum_{j=k}^{2k-1}(2j+1)-(2k+1)+(4k+1)+(4k+3)\;,$$
and since $(4k+1)+(4k+3)-(2k+1)=6k+3$, we have
$$\sum_{j=k+1}^{2k+1}(2j+1)=\sum_{j=k}^{2k-1}(2j+1)+6k+3\;,$$
The induction hypothesis is that $(2)$ is $3n^2$, so this simplifies further to
$$\sum_{j=k+1}^{2k+1}(2j+1)=3n^2+6k+3=3(n^2+2n+1)=3(n+1)^2\;.\;,$$
A: The boundaries of the summation were 2k+1 and k+1, so by extracting the last 2 terms and adding a new first term, the boundaries change to k and 2k-1. He then subs in the assumption, expands/simplifies then factor out the three to make a perfect square
