Sprawdzamy prawdziwość równości dla n = 1:
[tex]\text{L}=\sum\limits_{k=1}^1\dfrac{1}{k(k+1)(k+2)}=\dfrac{1}{1\cdot2\cdot3}=\dfrac{1}{6}\\\\\\\text{P}=\dfrac{1}{2}\Big(\dfrac{1}{2}-\dfrac{1}{(1+1)(1+2)}\Big)=\dfrac{1}{2}\Big(\dfrac{1}{2}-\dfrac{1}{6}\Big)=\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{6}\\\\\\\text{L}=\text{P}[/tex]
Zakładamy, że:
[tex]\sum\limits_{k=1}^n\dfrac{1}{k(k+1)(k+2)}=\dfrac{1}{2}\Big(\dfrac{1}{2}-\dfrac{1}{(n+1)(n+2)}\Big)[/tex]
Chcemy pokazać, że:
[tex]\sum\limits_{k=1}^{n+1}\dfrac{1}{k(k+1)(k+2)}=\dfrac{1}{2}\Big(\dfrac{1}{2}-\dfrac{1}{(n+2)(n+3)}\Big)[/tex]
Mamy:
[tex]\text{L}=\sum\limits_{k=1}^{n+1}\dfrac{1}{k(k+1)(k+2)}=\sum\limits_{k=1}^n\dfrac{1}{k(k+1)(k+2)}+\dfrac{1}{(n+1)(n+2)(n+3)}=\\\\\\\underset{ind}{\overset{zal}{=}}\dfrac{1}{2}\Big(\dfrac{1}{2}-\dfrac{1}{(n+1)(n+2)}\Big)+\dfrac{1}{(n+1)(n+2)(n+3)}=\\\\\\=\dfrac{1}{4}+\dfrac{-1}{2(n+1)(n+2)}+\dfrac{1}{(n+1)(n+2)(n+3)}=\\\\\\=\dfrac{1}{4}+\dfrac{-(n+3)}{2(n+1)(n+2)(n+3)}+\dfrac{2}{2(n+1)(n+2)(n+3)}=\\\\\\=\dfrac{1}{4}+\dfrac{-(n+1)}{2(n+1)(n+2)(n+3)}=\dfrac{1}{4}+\dfrac{-1}{2(n+2)(n+3)}=[/tex]
[tex]=\dfrac{1}{2}\Big(\dfrac{1}{2}-\dfrac{1}{(n+2)(n+3)}\Big)=\text{P}[/tex]
