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

Question

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?

Answer 1

When you plug in $n=k+1$ you have that $$5+10+\cdots+5k+5(k+1)\not=5+10+\cdots+5k+5k+1$$

Answer 2

Since you have your answer now! Let's see it with a different approach

$$S=\color{red}{5}+\color{blue}{10}+\cdots+\color{green}{5(n-1)}+\color{orange}{5n}$$ $$S=\color{red}{5n}+\color{blue}{5(n-1)}+\cdots+\color{green}{10}+\cdots+\color{orange}{5}$$

$$2S=\underbrace{\color{red}{5(n+1)}+\color{blue}{5(n+1)}+\cdots+\color{green}{5(n+1)}+\color{orange}{5(n+1)}}_{\text{n terms}}$$

$$S=\frac{5n(n+1)}{2}$$

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\large\tt\mbox{A different approach}.\quad}$ 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, $$ \color{#66f}{\large\sum_{k\ =\ 1}^{n}k} =0 + \half\,n + \half\,n^{2} = \color{#66f}{\large{n\pars{n + 1} \over 2}} $$