Sum of Special Sequences
Calculates the sum of the selected special sequence up to the nth term.
| Sum of a Sequence |
| \(\normalsize (1)\ {\large\displaystyle \sum_{\small k=1}^{\small n}}k ={\large\frac{n(n+1)}{2}}\\ (2)\ {\large\displaystyle \sum_{\small k=1}^{\small n}}k^2 ={\large\frac{n(n+1)(2n+1)}{6}}\\ (3)\ {\large\displaystyle \sum_{\small k=1}^{\small n}}k^3 ={\large\frac{n^2(n+1)^2}{4}}\\ (4)\ {\large\displaystyle \sum_{\small k=1}^{\small n}}k(k+1) ={\large\frac{n(n+1)(n+2)}{3}}\\ (5)\ {\large\displaystyle \sum_{\small k=1}^{\small n}\frac{1}{k(k+1)}} ={\large\frac{n}{n+1}}\\ (6)\ {\large\displaystyle \sum_{\small k=1}^{\small n}}k(k+1)(k+2) ={\large\frac{n(n+1)(n+2)(n+3)}{4}}\\ (7)\ {\large\displaystyle \sum_{\small k=1}^{\small n}\frac{1}{k(k+1)(k+2)}} ={\large\frac{n(n+3)}{4(n+1)(n+2)}}\\ \) |
No customer feedback for this calculation yet.