Proving that $\sum_{i=2}^n(5i-4)=\frac{n(5n-3)-2}{2}$ for all $n\geq 1$ by mathematical induction I have this question:
Show, using mathematical induction, that for all natural numbers $n$,
$$6 + 11 + 16 + 21 + \cdots + (5n-4) = \frac{n(5n-3)-2}{2}$$

I am confused in that that question states for all $natural$ $numbers$... 
yet if you take $1$ (for example as the base case) which is a $natural$ $number$
it does not work: 
LHS: $(5n-4) = (5(1)-4) = 1$
RHS:  $\frac{n(5n-3)-2}{2} = \frac{1(5(1)-3)-2}{2} = 0$
Am I understaning/doing something wrong? 
Should I just take $S(2)$ as the base step?
How would I go about solving this problem?
 A: As the comments have indicated, the equation is actually valid for $n = 1$, after a fashion.  It is clearer when written as a summation:
$$
\sum_{k=2}^n (5k-4) = \frac{n(5n-3)-2}{2}
$$
Spelled out in words, the above equation reads, "The sum of $(5k-4)$, from $k = 2$ to $n$, equals $\ldots$"  When $n = 1$, the summation on the left is empty, and an empty sum is by definition equal to zero (which equals the fraction on the right).  For a more familiar identity, we might write
$$
\sum_{k=1}^n k = 1+2+3+\cdots+n = \frac{n(n+1)}{2}
$$
When $n = 0$, the summation on the left is empty, so the sum is zero (which, again, equals the fraction on the right).  The confusing part of your question is that the summand (the $(5k-4)$ you're adding up) is not actually zero for $n = 1$, so it is easy to make the mistake of using it in the sum.  But in fact the limits of the sum dictate that it not be included.
Anyway, to prove by induction: The above will serve as the basis case.  The induction step is to assume that
$$
\sum_{k=2}^{n-1} (5k-4) = \frac{(n-1)[5(n-1)-3]-2}{2}
$$
and show that adding $5n-4$ to both sides of the equation yields
$$
\sum_{k=2}^n (5k-4) = \frac{n(5n-3)-2}{2}
$$
Can you take it from here?
A: What Daniel Fischer, Brian Tung, Michael Hardy, and myself now are referring to is that the empty sum is taken to be zero; on the other hand, the empty product is taken to be one. More concretely, 
$$
\sum_{i=r}^k\Omega_i=0\quad\text{when}\quad k<r\quad\text{and}\quad\prod_{i=r}^k\Omega_i=1\quad\text{when}\quad k<r.
$$
Now, consider Brian's comment; you are trying to prove that
$$
\sum_{i=2}^n(5i-4)=\frac{n(5n-3)-2}{2}
$$
for all natural numbers $n$ (i.e., $n\geq 1$, where $n\in\mathbb{Z}$). Let's do this, now that we have addressed the empty sum issue.

Claim: For all $n\geq 1$, where $n\in\mathbb{Z}$, we have that $\sum_{i=2}^n(5i-4)=\frac{n(5n-3)-2}{2}$.
Proof. For $n\geq 1$, let $S(n)$ denote the statement
$$
S(n) : \sum_{i=2}^n(5i-4)=\frac{n(5n-3)-2}{2}.
$$
Base case ($n=1$): $S(1)$ says that $\sum_{i=\color{purple}{2}}^\color{red}{1}(5i-4)=\underbrace{0}_{\text{since $\color{red}{1}<\color{purple}{2}$}}=\frac{(5-3)-2}{2}$, and this is true.
Inductive step $S(k)\to S(k+1)$: Fix some $k\geq 1$, and suppose that 
$$
S(k) : \color{blue}{\sum_{i=2}^k(5i-4)}=\color{blue}{\frac{k(5k-3)-2}{2}}
$$
holds. To be shown is that $S(k+1)$ follows where
$$
S(k+1) : \color{green}{\sum_{i=2}^{k+1}(5i-4)}=\color{green}{\frac{(k+1)(5k+2)-2}{2}}.
$$
Starting with the left-hand side of $S(k+1)$,
\begin{align}
\color{green}{\sum_{i=2}^{k+1}(5i-4)} &= \color{blue}{\sum_{i=2}^k(5i-4)}+[5(k+1)-4]\tag{by defn. of $\Sigma$}\\[0.5em]
&= \color{blue}{\frac{k(5k-3)-2}{2}}+5k+1\tag{by $S(k)$, the ind. hyp.}\\[1em]
&= \frac{5k^2-3k-2+10k+2}{2}\tag{common denom}\\[1em]
&= \frac{5k^2+7k+2-2}{2}\tag{simplify / rearrange}\\[1em]
&= \color{green}{\frac{(k+1)(5k+2)-2}{2}},\tag{factor}
\end{align}
we end up at the right-hand side of $S(k+1)$, completing the inductive step. 
Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$
