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Sa se calculeze:
\( S_m=\sum_{n=1}^m {\[\sqrt{n^2+10n}\]} \)

Concursul Gh.Titeica

Sa observam ca
\( n\ge8\Longrightarrow\left(n+4\right)^{2}\le n^{2}+10n<\left(n+5\right)^{2}\Longrightarrow n+4\le\sqrt{n^{2}+10n}<n+5\Longrightarrow\left[\sqrt{n^{2}+10n}\right]=n+4 \).
Atunci \( S_{m}=\sum_{n=1}^{7}\left[\sqrt{n^{2}+10n}\right]+\sum_{n=8}^{m}\left[\sqrt{n^{2}+10n}\right] \).

Dar cum \( \sum_{n=1}^{7}\left[\sqrt{n^{2}+10n}\right]=3+4+6+7+8+9+10=47 \) si \( \sum_{n=8}^{m}\left[\sqrt{n^{2}+10n}\right]=\sum_{8}^{m}n+4=\sum_{12}^{m+4}a=\sum_{1}^{m+4}a-\sum_{1}^{11}a=\frac{\left(m+4\right)\left(m+5\right)}{2}-6\cdot11 \) deducem ca \( S_{m}=\frac{\left(m+4\right)\left(m+5\right)}{2}-19 \) este suma cautata.