Somma telescopica
\[\begin{eqnarray}\nonumber 3\sum_{k=1}^{n}k^2 &=& \sum_{k=1}^{n}(k + 1)^3 - \sum_{k=1}^{n}k^3 - 3\sum_{k=1}^{n}k - n \cr 3\sum_{k=1}^{n}k^2 &=& (n+1)(n+1)^2 - 1 -\frac{3}{2}n(n+1) - n \cr &=& (n+1)\left[(n+1)^2
-\frac{3}{2}n -1\right]\cr &=& \cdots \cr &=& \frac{1}{2}n(n+1)(2n+1) \end{eqnarray}\] Infine abbiamo: \[ \color{brown}{\begin{eqnarray}\nonumber \boxed{\sum_{k=1}^{n}k^2 = \frac{1}{6}n(n+1)(2n+1)} \end{eqnarray}} \] |
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Digressioni Sviluppiamo \( \sum_{k=1}^{n} r_k^2 \) osservando che \[ (k + 1)^3 - k^3 = 3k^2 + 3k + 1 \] \[\sum_{k=1}^{n}(k+1)^3 - k^3 = (n+1)^3 - 1 \] Somma telescopica \[ \begin{eqnarray}\nonumber 1^3 &=& 1^3\cr 2^3 &=& (1 +1)^3 = 1^3 + 3 + 3 + 1^3 = 8\cr 3^3 &=& (2+1)^3 = 2^3 +3\cdot 2^2 + 3\cdot 2 + 1^3 \cr &=& 8+12 + 6 + 1 = 27\cr &\cdots&\cr (k+1)^3 &=& k^3+3k^2 + 3k +1 \end{eqnarray} \] da cui: \[ 3k^2 = (k+1)^3 - 3k - 1 \] |
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