Let $P_n$ be a proposition, so that $$P_n: 3+11+...+(8n-5)=4n^2-n$$ $P_1:3=4\cdot1^2-1$, $P_2:3+11=4\cdot2^2-2$ etc.
When proving $P_n\to P_{n+1}$, is it enough to show in the second induction step that, by subtracting from both sides we have $$4(n+1)^2-(n+1)=4n^2-n+(8(n+1)-5)\leftrightarrow0=0$$ and therefore $P_n\to P_{n+1}$. Or do I must rewrite $4n^2-n+(8(n+1)-5)$ to $4(n+1)^2-(n+1)$?
I would rather make the use of induction extremely clear.
First of all, I would write $P_n: 3+11+...+(8n-5)=4n^2-n $ as $P_n: \sum_{k=1}^n (8n-5) =4n^2-n $. This makes the start and end terms of the sum clear and reduces the chance of error.
Then I would show the step from $P_n$ to $P_{n+1}$ like this:
$\begin{array}\\ \sum_{k=1}^{n+1} (8n-5) &=\sum_{k=1}^{n} (8n-5)+8(n+1)-5 \qquad\text{split off the last term}\\ &=4n^2-n+8(n+1)-5 \qquad\text{use the induction hypothesis}\\ &=4n^2+7n+3 \qquad\text{algebra}\\ &=4(n+1)^2-(n+1) \qquad\text{more algebra to get the right side of }P_{n+1}\\ \end{array} $
which is $P_{n+1}$, and we are done.
More generally, suppose we are given $a$ and $b$, and we want to find $u, v, $ and $w$ such that $P_n: \sum_{k=1}^n (an+b) =un^2+vn+w $ is true.
Arguing exactly as above,
$\begin{array}\\ \sum_{k=1}^{n+1} (an+b) &=\sum_{k=1}^{n} (an+b)+a(n+1)+b \qquad\text{split off the last term}\\ &=un^2+vn+w+a(n+1)+b \qquad\text{use the induction hypothesis}\\ &=un^2+(v+a)n+w+a+b \qquad\text{algebra}\\ \end{array} $
$P_{n+1}$ is $\sum_{k=1}^{n+1} (an+b) =u(n+1)^2+v(n+1)+w $, so we want
$un^2+(v+a)n+w+a+b\\ =u(n+1)^2+v(n+1)+w\\ =un^2+(2u+v)n+u+v+w. $
Equating coefficients, $v+a = 2u+v$ and $w+a+b = u+v+w$.
From the first, $u = a/2$.
From the second, $a+b = u+v =a/2+v$ so $v = a/2+b$.
For your problem, $a=8, b=-5$, so $u = 4, v=8/2-5 =-1 $.
Note that $w$ does not seem to be determined. For this, we need the initial value; either $n=0$ or $n=1$ will work.
Using $n=0$, the sum is empty, so we want $0 = w$.
Using $k=1$, $P_1$ is $a+b = u+v+w$ or $w =a+b-u-v =a+b-a/2-(a/2+b) = 0 $.
It is comforting that both cases lead to the same result.
Therefore we have shown that $\sum_{k=1}^n (ak+b) =(a/2)n^2+(a/2+b)n $.
This type of argument enables us to both discover a result and prove it.
In some problems, this technique allows us to show that a particular form of a summation does not exist, because the assumption that the form exists leads to a contradiction.