Formula

\[\sum_{i=1}^n i=\frac{n(n+1)}{2}\\ \sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}\\ \sum_{i=1}^n i^3=\frac{n^2(n+1)^2}{4}\\ \sum_{i=1}^n i^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}\\ \log_{a}b=\frac{\log_{c}b}{\log_{c}a} \]