Respuesta :

WojtekR
From "Series 1" we know, that:

[tex]1+2+3+\ldots+n=\dfrac{n(n+1)}{2}[/tex]

and from "Series 2":

[tex]1^3+2^3+3^3+\ldots+n^3=(1+2+3+\ldots+n)^2[/tex]

c)

[tex]3^3+6^3+9^3+\ldots+30^3=(1\cdot3)^3+(2\cdot3)^3+(3\cdot3)^3+\ldots+(10\cdot3)^3=\\\\=1^3\cdot3^3+2^3\cdot3^3+3^3\cdot3^3+\ldots+10^3\cdot3^3=\\\\=(1^3+2^3+3^3+\ldots+10^3)\cdot3^3\,\stackrel{\text{Series 2}}{=}\,(1+2+3+\ldots+10)^2\cdot27=\\\\\stackrel{\text{Series1}}{=}\,\left(\dfrac{10\cdot11}{2}\right)^2\cdot27=55^2\cdot27=3025\cdot27=\boxed{81675}[/tex]

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