Q1279 (Q1279): Difference between revisions

@@ Property / relevant answers @@
+https://math.stackexchange.com/a/3189081
@@ Property / relevant answers: https://math.stackexchange.com/a/3189081 / rank @@
+Normal rank
+Defining formula:  $1^{2}+\ldots +p^{2}+(p+1)^{2}={\frac {p(p+1)(2p+1)}{6}}+(p+1)^{2}={\frac {p+1}{6}}\left[p(2p+1)+6(p+1)\right]={\frac {p+1}{6}}\left[2p^{2}+8p+6\right]={\frac {(p+1)(p+2)(2p+3)}{6}}$ 
1^2+\ldots+p^2+(p+1)^2 = \frac{p(p+1)(2p+1)}{6} + (p+1)^2 = \frac{p+1}{6} \left[p(2p+1)+6(p+1)\right] = \frac{p+1}{6} \left[2p^2 + 8p+6\right] = \frac{(p+1)(p+2)(2p+3)}{6}

Revision as of 15:07, 13 May 2020