suma $L$ de la $1$ la $N-1$ de $(N–L) × (N–L) × L =$ suma cu $L$ de la $1$ la $N–1$ de $LN^2^–2 NL^2^ + L^3^ = N^2^ ×$ suma $L$ de la $1$ la $N-1-2N$ suma $L$ de la $1$ la $N–1$ de $L^2^ +$ suma de $L$ de la $1$ la $N–1$ de $L^3^ = N^2^ ×  N(N-1)/2-2N n(n-1)(2n-1)/6 + [n(n-1)/2]^2^$.
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\[ \sum_{L=2}^N {(N-L+1)(N-L+1)(L-1)} \] =
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= \[\sum_{L=1}^{N-1} {(N-1)(N-1)L} \] =
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 = \[\sum_{L=1}^{N-1} {LN^2 - 2NL^2 + L^3}\] =
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= \[N^2\sum_{L=1}^{N-1} {L} - 2N\sum_{L=1}^{N-1} {L^2} + \sum_{L=1}^{N-1} {L^3} \] =
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 = \[n^2\fraq {N(N-1)}{2} - 2n\fraq{N(N-1)(2N-1)}{6} + \left[\fraq{N(N-1)}{2}\rigt]^2 .\]

\begin{array} {lcl} \[ \sum_{L=2}^N {(N-L+1)(N-L+1)(L-1)} \] & = & \[\sum_{L=1}^{N-1} {(N-1)(N-1)L} \] \\ & = & \[\sum_{L=1}^{N-1} {LN^2 - 2NL^2 + L^3}\] \\ & = & \[N^2\sum_{L=1}^{N-1} {L} - 2N\sum_{L=1}^{N-1} {L^2} + \sum_{L=1}^{N-1} {L^3} \] \\ & = & \[n^2\fraq {N(N-1)}{2} - 2n\fraq{N(N-1)(2N-1)}{6} + \left[\fraq{N(N-1)}{2}\rigt]^2 .\] \end{array}

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Am folosit formulele cunoscute din manualul de clasa a 10-a: