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mathmath333
 one year ago
Solve sum(k=1)^m [ k*(nm+k) ]
mathmath333
 one year ago
Solve sum(k=1)^m [ k*(nm+k) ]

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & \sum_{k=1}^m k \cdot (nm+k) \hspace{.33em}\\~\\ \end{align}}\)

***[ISURU]***
 one year ago
Best ResponseYou've already chosen the best response.2\[=\sum_{k=1}^{m}k.(n  m ) + \sum_{k=1}^{m}k.k\]\[=(n m ) \sum_{k=1}^{m}k + \sum_{k=1}^{m}k^2 \]\[=( n m) (\frac{ m }{ 2 } ) ( 1 + m ) + \frac{ m }{ 6 } (n + 1 )(2n + 1)\]

***[ISURU]***
 one year ago
Best ResponseYou've already chosen the best response.2because \[\sum_{k=1}^{m} k = \frac{ m }{ 2 }( m + 1)\] and\[\sum_{k=1}^{m}k^2 = \frac{ m }{ 6 }( m +1 ) ( 2m +1)\]

***[ISURU]***
 one year ago
Best ResponseYou've already chosen the best response.2srry... the last answer should be... \[=( n  m) ( \frac{ m }{ 2} ) ( m + 1) + \frac{ m }{ 6 } ( m +1) ( 2m + 1)\]
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