Vollständige Induktion!

Vollständige Induktion!

$\sum \limits_{k=0}^{n-1}(n+k)(n-k) = \dfrac{n(n+1)(4n-1)}{6}$

Führen sie einen Beweis mittels vollständiger Induktion durch.

Induktionsannahme:
$\sum \limits_{k=0}^{n-1}(n+k)(n-k) = \dfrac{n(n+1)(4n-1)}{6}$

Induktionsannahme für $n=1$:

$LS = \sum \limits_{k=0}^{1-1}(1-0)(1+0)=1.$

$RS = \dfrac{1(1+1)(4*1-1)}{6} = 1$

Induktionsschritt für $n+1$:

$\sum \limits_{k=0}^{(n+1)-1}\Big((n+1)+k \Big) \Big((n+1)-k \Big) = \dfrac{(n+1)\Big((n+1)+1\Big)\Big(4(n+1)-1\Big)} {6} \\ \sum \limits_{k=0}^{n}\Big((n+1)+k \Big) \Big((n+1)-k \Big) = \dfrac{(n+1)(n+2)(4n+3)} {6}$

$\begin{array}{|rcll|} \hline LS &=& \sum \limits_{k=0}^{n}\Big((n+1)+k \Big) \Big((n+1)-k \Big) \\ &=& \sum \limits_{k=0}^{n-1}\Big((n+1)+k \Big) \Big((n+1)-k \Big) + \overbrace{\Big((n+1)+n \Big) \Big( (n+1)-n \Big)}^{k=n} \\ &=& \sum \limits_{k=0}^{n-1}\Big((n+1)+k \Big) \Big((n+1)-k \Big) + (2n+1)*1 \\ &=& \sum \limits_{k=0}^{n-1}\Big((n+1)+k \Big) \Big((n+1)-k \Big) + (2n+1) \\ &=& \sum \limits_{k=0}^{n-1}\Big((n+1)^2-k^2 \Big)+ (2n+1) \\ &=& \sum \limits_{k=0}^{n-1}\Big(n^2+2n+1-k^2 \Big)+ (2n+1) \\ &=& \sum \limits_{k=0}^{n-1}\Big(n^2-k^2+2n+1 \Big)+ (2n+1) \\ &=& \sum \limits_{k=0}^{n-1}\Big(n^2-k^2 \Big)+\sum \limits_{k=0}^{n-1}\Big(2n+1 \Big)+ (2n+1) \\ &=& \underbrace{\sum \limits_{k=0}^{n-1}\Big((n+k)(n-k)\Big)}_{=\frac{n(n+1)(4n-1)}{6}}+\sum \limits_{k=0}^{n-1}\Big(2n+1 \Big)+ (2n+1) \\ &=& \dfrac{n(n+1)(4n-1)}{6}+\underbrace{\sum \limits_{k=0}^{n-1}\Big(2n+1 \Big)}_{=(2n+1)n}+ (2n+1) \\ &=& \dfrac{n(n+1)(4n-1)}{6}+ (2n+1)n+ (2n+1) \\\\ &=& \dfrac{n(n+1)(4n-1)}{6}+ (2n+1)(n+1) \\\\ &=& \dfrac{n(n+1)(4n-1)+6(2n+1)(n+1)}{6} \\\\ &=& \dfrac{(n+1)\Big( n(4n-1)+6(2n+1) \Big) }{6} \\\\ &=& \dfrac{(n+1)( 4n^2-n+12n+6) }{6} \\\\ \mathbf{LS} &=& \mathbf{\dfrac{(n+1)(4n^2+11n+6)} {6}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline RS &=& \dfrac{(n+1)(n+2)(4n+3)} {6} \\\\ &=& \dfrac{(n+1)(4n^2+3n+8n+6)} {6} \\\\ \mathbf{RS} &=& \mathbf{\dfrac{(n+1)(4n^2+11n+6)} {6}} \\ \hline \end{array}$