$$1+2+3+...+n=\frac{n(n+1)}{2}$$
$$1+3+5+...+(2n-1)=n^2$$
$$2+4+6+...+2n=n(n+1) $$
$$1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$
$$1^2+3^2+5^2+...+(2n-1)^2=\frac{n(4n^2-1)}{3} $$
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...+\frac{1}{n(n+1)}=\frac{n}{n=1}$$
$$\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+...+\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{(n+1)(n+2)} \right]$$
2010-12-07 • Просмотров [ 2524 ]
