%&latex
\section{Proposizioni varie}
\begin{prop}\label{sommapriminnumeri}
	\eal{&\sum_{i=1}^n i =\frac{n(n+1)}{2}}
\end{prop}
\begin{proof}
	Supponiamo dapprima $n$ dispari:
\eal{&\sum_{i=1}^n i =1+2+3+\dots+(n-3)+ (n-2)+(n-1)+n=
\Big(1+\big(2+(3+\dots+ n-3)+n-2\big)+n-1\Big)+n=\\
&=
(\ubrace{n+n+n+\dots+n}{(n-1)/2\T{
volte}})+n=\frac{n-1}{2}n+n=\frac{n(n+1)}{2}
	}
Se $n$ e' pari:
\eal{&\sum_{i=1}^n i =n+\sum_{i=1}^{n-1} i \ubrace{=}{n-1\T{ dispari}} 
n+\frac{(n-1)n}{2}=\frac{2n+n^2-n}{2}\frac{n(n+1)}{2}
}
\end{proof}

\begin{prop}\label{sommaquadrati}
	Somma quadrati:
	\eal{&\sumin i^2=\frac{2n^3+3n^2+n}{6}
	}
\end{prop}
\begin{proof}
	\def\D{{\Delta}}
	Data una successione $\{a_i\}$, poniamo:
\eal{&\D a_i = a_{i+1}-a_i}
Si ha
	\eal{&\D i^3 =(i+1)^3-i^3=1+3i^2+3i\\
	&3i^2=\D i^3 -3i-1\\
	&3\sumin i^2 = \sumin \pton{\D i^3 -3i-1} = \\
	&\qq \sumin \D i^3 = \sumin (i+1)^3-i^3 = 2^3-1^3 + 3^3-2^3 + \dots +
	(n+1)^3-n^3 = (n+1)^3-1^3\\
	&= (n+1)^3-1^3 -3\sumin i - \sumin 1
	\ubrace{=}{\see{sommapriminnumeri}} (n+1)^3-1^3 
	-3\frac{n(n+1)}{2} - n\\
	&\sumin i^2 = \frac{
	 2(n+1)^3-2
	         -3n(n+1) - 2n}{6}
		 =
		 \frac{2n^3+2+6n^2+6n-2-3n^2-3n-2n}{6}=
		 \frac{2n^3+3n^2+n}{6}
	}
\end{proof}