Proving that $3+11 +19 + \cdots + (8n-5)= 4n^2 -n$ by induction 
$ 3+11 +19 + \cdot \cdot \cdot + (8n-5)= 4n^2 -n.$

I rewrote the problem in sigma notation. Would this change the answer?
$$ \sum_{i=1}^n (8n-5) = 4n^2-n.$$ 
$\color{red}{Proof :}$
(i) $\;$ The statement is true for $n=1$ because
$\left( 8(1)-5) = 4(1)^2-1 \right) \Rightarrow 3=3$ 
(ii) $\;$ Let $n =k $ $\Rightarrow \; \text{Induction Hypothesis} $$$\sum_{i=1}^{k} (8k-5) = 4(k)^2-k$$
$\text{We need to show :}$ $ n= k+1$
$$\sum_{i=1}^{k+1} \left( 8(k+1)-5 \right) = 4(k+1)^2-(k+1)$$
$= \left( 4(k)^2-k \right) + \left( 8(k+1)-5 \right) \Rightarrow 4k^2+7k+3$
Thus the statement is true for $n+1$ 
(iii) By the principle of PMI, statement is true for every $n \in  ℕ$.
Yet here is my question. If I write this question in sigma notation would the answer still be correct. Or do I have to write this answer without it. Would this notation be fine? 
 A: As Andre Nicolas points out, your primary issue occurs in how you write your summation. If you do it correctly, then your work should look, more or less, as follows (side bar comments removed...see if you can follow one line to the next):
\begin{align}
\sum_{i=1}^{k+1}(8i-5)&= \sum_{i=1}^k(8i-5)+8(k+1)-5\\[1em]
&= (4k^2-k)+8(k+1)-5\\[1em]
&= 4k^2+7k+3\\[1em]
&= 4k^2+8k+4-k-1\\[1em]
&= 4(k+1)^2-(k+1).
\end{align}
A: Let $S(n)$ be the statement: $3+11+19+\cdots+(8n-5)=4n^{2}-n$
Basis step: $S(1)$:
LHS: $\big(8(1)-5\big)=3$
RHS: $4(1)^{2}-(1)=3$
$\hspace{55 mm}$ LHS $=$ RHS $\hspace{1 mm}$ (verified.)
Inductive step:
Assume $S(k)$ is true, i.e. assume that $3+11+19+\cdots+(8k-5)=4k^{2}-k$
For $S(k+1)$:
LHS: $\hspace{7 mm}\underline{3+11+19+\cdots+(8k-5)}+\big(8(k+1)-5\big)$
$\hspace{12 mm}=4k^{2}-k+8k+8-5$
$\hspace{12 mm}=4k^{2}+7k+3$
$\hspace{12 mm}=4k^{2}+4k+3k+3$
$\hspace{12 mm}=4k\hspace{1 mm}(k+1)+3\hspace{1 mm}(k+1)$
$\hspace{12 mm}=(k+1)(4k+3)$
RHS: $\hspace{6 mm}4\hspace{1 mm}(k+1)^{2}-(k+1)$
$\hspace{12 mm}=4\hspace{1 mm}(k^{2}+2k+1)-k-1$
$\hspace{12 mm}=4k^{2}+8k+4-k-1$
$\hspace{12 mm}=4k^{2}+7k+3$
$\hspace{12 mm}=4k^{2}+4k+3k+3$
$\hspace{12 mm}=4k\hspace{1 mm}(k+1)+3\hspace{1 mm}(k+1)$
$\hspace{12 mm}=(k+1)(4k+3)$
$\hspace{55 mm}$ LHS $=$ RHS $\hspace{1 mm}$ (verified.)
So, $S(k+1)$ is true whenever $S(k)$ is true.
Therefore, $3+11+19+\cdots+(8n-5)=4n^{2}-n$.
