Q1279 (Q1279): Difference between revisions
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Created claim: relevant answers (P14): https://math.stackexchange.com/a/3189081 |
Changed claim: relevant answers (P14): 3189081 |
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| Property / relevant answers | Property / relevant answers | ||
| Property / relevant answers: https://math.stackexchange.com/a/3189081 / qualifier | |||
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} | |||
![{\displaystyle 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/833b59bd657ff62c2b6c76586804439d904e1256)
Latest revision as of 13:46, 20 May 2020
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A.71
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