Weak principle of induction for $5+10+15+\ldots+5n= \frac{5n(n+1)}{2}$ 
Show that
$$5+10+15+\ldots+5n= \frac{5n(n+1)}{2}$$

Proving the base case $n(1)$:
$5(1)= \frac{5(1)(1+1)}{2}$
$5 = \frac{5(2)}{2}$
$5 = 5$
Induction hypothesis:
$n = k$
$5+10+15+\ldots+5k = \frac{5k(k+1)}{2}$
Induction step (adding $k+1$):
$5+10+15+\ldots+5k+5k+1 = \frac{5k+1(k+1+1)}{2}$
Substituting $\frac{5k(k+1)}{2}$ for $5k$:
$\frac{5k(k+1)}{2}+5k+1 = \frac{(5k+1)(k+1+1)}{2}$
Simplifying:
$\frac{5k(k+1)+2(5k+1)}{2} = \frac{(5k+1)(k+2)}{2}$
$\frac{5k^2+5k+10k+2}{2} = \frac{5k^2+10k+k+2}{2}$
These aren't equal, so what did I do wrong here?
 A: 
Since you have your answer now! Let's see it with a different approach

$$S=\color{red}{5}+\color{blue}{10}+\cdots+\color{green}{5(n-1)}+\color{orange}{5n}$$
$$S=\color{red}{5n}+\color{blue}{5(n-1)}+\cdots+\color{green}{10}+\cdots+\color{orange}{5}$$
$$2S=\underbrace{\color{red}{5(n+1)}+\color{blue}{5(n+1)}+\cdots+\color{green}{5(n+1)}+\color{orange}{5(n+1)}}_{\text{n terms}}$$
$$S=\frac{5n(n+1)}{2}$$
A: When you plug in $n=k+1$ you have that
$$5+10+\cdots+5k+5(k+1)\not=5+10+\cdots+5k+5k+1$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{\large\tt\mbox{A different approach}.\quad}$
Lets $\ds{\sum_{k\ =\ 1}^{n}k=a_{0} + a_{1}\,n + a_{2}\,n^{2}}$

\begin{align}
\color{#66f}{\large a_{2}}&=\lim_{n\ \to\ \infty}{1 \over n^{2}}\sum_{k\ =\ 1}^{n}k
=\lim_{n\ \to\ \infty}
{\sum_{k\ =\ 1}^{n + 1}k - \sum_{k\ =\ 1}^{n}k \over \pars{n + 1}^{2} - n^{2}}
=\lim_{n\ \to\ \infty}{n + 1 \over 2n + 1} =\color{#66f}{\large\half}
\end{align}

\begin{align}
\color{#66f}{\large a_{1}}&
=\lim_{n\ \to\ \infty}{1 \over n}\pars{\sum_{k\ =\ 1}^{n}k - a_{2}\,n^{2}}
=\lim_{n\ \to\ \infty}{1 \over n}\pars{\sum_{k\ =\ 1}^{n}k - \half\,n^{2}}
\\[5mm]&=\lim_{n\ \to\ \infty}
{\sum_{k\ =\ 1}^{n + 1}k - \pars{n + 1}^{2}/2 - \sum_{k\ =\ 1}^{n}k + n^{2}/2\over
\pars{n + 1} - n}
=\lim_{n\ \to\ \infty}\pars{n + 1 - n - \half} =\color{#66f}{\large\half}
\end{align}

$$
1 = a_{0} + a_{1} + a_{2} = a_{0} + 1\quad\imp\quad
\color{#66f}{\large a_{0}} = \color{#66f}{\large 0}
$$

Then,
$$
\color{#66f}{\large\sum_{k\ =\ 1}^{n}k}
=0 + \half\,n + \half\,n^{2} = \color{#66f}{\large{n\pars{n + 1} \over 2}}
$$
