$ \sum{ k^3}\;的公式 $

$ Pf： $

$ 考慮：\;(k+1)^4-k^4=4k^3+6k^2+4k+1$

$ \sum_{k=1}^{n}{\left [ (k+1)^4-k^4 \right ] }=\sum_{k=1}^{n}{ \left (4k^3+ 6k^2+4k+1 \right ) }$

$⇒ (2^4-1^4)+(3^4-2^4)+...+[n^4-(n-1)^4]+[(n+1)^4-n^4]$
$\;\;=4\sum_{k=1}^{n}{k^3 }+6\sum_{k=1}^{n}{k^2 } +4\sum_{k=1}^{n}{k }+\sum_{k=1}^{n}{1 }$

$⇒ (n+1)^4-1=4\sum_{k=1}^{n}{ k^3}+6\times\frac{n(n+1)(2n+1)}{6}+4\times\frac{n(n+1)}{2}+n$

$⇒n^4+4 n^3+6n^2+4n=4\sum_{k=1}^{n}{ k^3}-n(n+1)(2n+1)-2\times n(n+1)+n$

$⇒4 \sum_{k=1}^{n}{k^3 } =(n^4+4n^3+6n^2+4n)-(2n^3+3n^2+n)-(2n^2+2n)-n$

$⇒\sum_{k=1}^{n}{ k^3}=\frac{n^4+2n^3+n^2}{4}=\left( \frac{n(n+1)}{2}  \right) ²$

n = 1 (slider from 1 to 50, step 1)
s = 0 (slider from 0 to 7, step 1)
A = (1.26, 5.82)
T1: "$ \sum_{k=1}^{n}{k^3 }=(\frac{n(n+1)}{2} )^2$
$\sum_{k=1}^{" + (LaTeX[n]) + "}{ k^3}=" + (LaTeX[If[n ≟ 1, "1^3", If[n ≟ 2, "1^3+2^3", If[n ≟ 3, "1^3+2^3+3^3", "1^3+2^3+..."]]]]) + "" + (LaTeX[If[n < 4, "", "+"]]) + "" + (LaTeX[If[n > 3, n * "^3"]]) + "$
$\;\;\;\;\;\;\;\;\;\;=\left(    \frac{" + (LaTeX[n]) + "×(" + (LaTeX[n]) + "+1)}{2}\right)^2$
$\;\;\;\;\;\;\;\;\;\;=\left(   \frac{" + (LaTeX[n]) + "×" + (LaTeX[n + 1]) + "}{2}\right)^2$
$\;\;\;\;\;\;\;\;\;\;=" + (LaTeX[(n^(2) * (n + 1)^(2)) / 4]) + "$"