Rozwiązanie:
Ustalmy ciąg arytmetyczny:
[tex]a_{n}=a_{1}+(n-1)r[/tex]
Wtedy mamy:
[tex]$S_{3n}=\frac{2a_{1}+(3n-1)r}{2} \cdot 3n[/tex]
[tex]$S_{2n}=\frac{2a_{1}+(2n-1)r}{2} \cdot 2n[/tex]
[tex]$S_{n}=\frac{2a_{1}+(n-1)r}{2} \cdot n[/tex]
Zatem mamy pokazać, że:
[tex]$\frac{2a_{1}+(3n-1)r}{2} \cdot 3n=3(\frac{2a_{1}+(2n-1)r}{2} \cdot 2n-\frac{2a_{1}+(n-1)r}{2} \cdot n)[/tex]
[tex]$\frac{6a_{1}n+9n^{2}r-3nr}{2} =3(2a_{1}n+2n^{2}r-nr-\frac{2a_{1}n+n^{2}r-nr}{2} )[/tex]
[tex]$3 \cdot \frac{2a_{1}n+3n^{2}r-nr}{2} =3 \cdot \frac{4a_{1}n+4n^{2}r-2nr-2a_{1}n-n^{2}r+nr}{2}[/tex]
[tex]$\frac{2a_{1}n+3n^{2}r-nr}{2}=\frac{2a_{1}n+3n^{2}r-nr}{2}[/tex]
[tex]L=P[/tex]
co kończy dowód.
