Utente:Mark16/Sandbox 3

• $\sum _{n=0}^{m}2^{n}=2^{m+1}-1$
• $\sum _{n=0}^{m}3^{n}={\frac {1}{2}}(3^{m+1}-1)$
• $\sum _{n=0}^{m}4^{n}={\frac {1}{3}}(4^{m+1}-1)$
• $\sum _{n=0}^{m}k^{n}={\frac {k^{m+1}-1}{k-1}}$ se k diverso da 0,1

• $\sum _{n=0}^{m}{\frac {1}{2^{n}}}=2-{\frac {1}{2^{m}}}$
• $\sum _{n=0}^{m}{\frac {1}{3^{n}}}={\frac {1}{2}}(3-{\frac {1}{3^{m}}})$
• $\sum _{n=0}^{m}{\frac {1}{4^{n}}}={\frac {1}{3}}(4-{\frac {1}{4^{m}}})$
• $\sum _{n=0}^{m}{\frac {1}{k^{n}}}={\frac {1}{k-1}}(k-{\frac {1}{k^{m}}})$ se k diverso da 0,1

• $\sum _{n=0}^{m}{\frac {1}{k^{n}}}={\frac {1}{k^{m}}}\sum _{n=0}^{m}k^{n}$