Publicado en 14 marzo, 2023 por Edison Del RosarioTabla de Fórmulas Matemáticas Referencia: Lathi Sec. B p54 B.8-3 Sumatorias ∑k=mnrk=rn+1−rmr−1 , r≠1 \sum_{k=m}^{n} r^k = \frac{r^{n+1}-r^m}{r-1} \texttt{ , } r\neq 1k=m∑nrk=r−1rn+1−rm , r≠1 ∑k=0nk=n(n+1)2 \sum_{k=0}^{n} k = \frac{n(n+1)}{2}k=0∑nk=2n(n+1) ∑k=0nk2=n(n+1)(2n+1)6 \sum_{k=0}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}k=0∑nk2=6n(n+1)(2n+1) ∑k=0nkrk=r+[n(r−1)−1]rn+1(r−1)2 , r≠1 \sum_{k=0}^{n} k r^k = \frac{r+[n(r-1)-1]r^{n+1}}{(r-1)^2} \texttt{ , } r\neq 1 k=0∑nkrk=(r−1)2r+[n(r−1)−1]rn+1 , r≠1 ∑k=0nk2rk=r[(1+r)(1−rn)−2n(1−r)rn−n2(1−r)2rn](1−r)3 , r≠1 \sum_{k=0}^{n} k^2 r^k = \frac{r[(1+r)(1-r^n)-2n(1-r)r^n - n^2(1-r)^2 r^n] }{(1-r)^3} \texttt{ , } r\neq 1 k=0∑nk2rk=(1−r)3r[(1+r)(1−rn)−2n(1−r)rn−n2(1−r)2rn] , r≠1
