Derive the formula for the sum of the first $n$ squares using derivatives and integrals

Question

I wanted to prove the formula for sum of squares without using induction and thought using derivatives would be the easiest approach $$f(n)=\sum_{k=1}^nk^2=\sum_{k=0}^{n-1}(n-k)^2\\f'(n)=2\sum_{k=0}^{n-1}(n-k)=2\sum_{k=1}^n k=n^2+n\\\int f'(n)dn=\frac{n^3}{3}+\frac{n^2}{2}+C$$ Now there is a term $\frac{n}{6}$ which is missing,why is this approach wrong?

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \sum_{k = 1}^{n}k^{2}&=\lim_{t\ \to\ 0}\partiald[2]{}{t}\sum_{k = 0}^{n}\expo{tk} =\lim_{t\ \to\ 0}\partiald[2]{}{t} \bracks{\color{#00f}{\expo{\pars{n + 1}t} - 1 \over \expo{t} - 1}} \end{align} $\ds{{\tt\mbox{We just need an expansion of the}}\ \color{#00f}{blue}\ \mbox{expression up to}\ t^{2}\ !!!}$.

Another approach. With $\ds{\verts{z} < 1}$: \begin{align} \sum_{n = 0}^{\infty}z^{n}\sum_{k = 1}^{n}k^{2}& =\sum_{k = 0}^{\infty}k^{2}\sum_{n = k}^{\infty}z^{n} =\sum_{k = 0}^{\infty}k^{2}{z^{k} \over 1 - z} ={1 \over 1 - z}\sum_{k = 0}^{\infty}k^{2}z^{k} ={1 \over 1 - z}\,{z\pars{1 + z} \over \pars{1 - z}^{3}} ={z\pars{1 + z} \over \pars{1 - z}^{4}} \\[3mm]&=z\pars{1 + z}\sum_{n = 0}^{\infty}{\mbox{}-4 \choose n}\pars{-1}^{n}z^{n} \\[3mm]&=\sum_{n = 0}^{\infty}{\mbox{}-4 \choose n}\pars{-1}^{n}z^{n + 1} +\sum_{n = 0}^{\infty}{\mbox{}-4 \choose n}\pars{-1}^{n}z^{n + 2} \\[3mm]&=\sum_{n = 1}^{\infty}{\mbox{}-4 \choose n - 1}\pars{-1}^{n - 1}z^{n} +\sum_{n = 2}^{\infty}{\mbox{}-4 \choose n - 2}\pars{-1}^{n - 2}z^{n} \\[3mm]&=1 + \sum_{n = 1}^{\infty}\bracks{% {\mbox{}-4 \choose n - 1}\pars{-1}^{n - 1} + {\mbox{}-4 \choose n - 2}\pars{-1}^{n}}z^{n} \end{align}

\begin{align} \sum_{k = 1}^{n}k^{2}& ={\mbox{}-4 \choose n - 1}\pars{-1}^{n - 1} +{\mbox{}-4 \choose n - 2}\pars{-1}^{n} \\[3mm]&=\pars{-1}^{n - 1}{4 + n - 1 - 1 \choose n - 1}\pars{-1}^{n - 1} +\pars{-1}^{n - 2}{4 + n - 2 - 1 \choose n - 2}\pars{-1}^{n} \\[3mm]&={n + 2 \choose n - 1} + {n + 1 \choose n - 2} ={n + 2 \choose 3} + {n + 1 \choose 3} \\[3mm]&={\pars{n + 2}\pars{n + 1}n \over 6} + {\pars{n + 1}n\pars{n - 1} \over 6} ={\pars{n + 1}n \over 6}\,\bracks{\pars{n + 2} + \pars{n - 1}} \\[3mm]&={n\pars{n + 1}\pars{2n + 1} \over 6} \end{align}