Question

Solve sum(k=1)^m [ k*(n-m+k) ]

Answer 1

\(\large \color{black}{\begin{align}
& \sum_{k=1}^m k \cdot (n-m+k) \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

\[=\sum_{k=1}^{m}k.(n - m ) + \sum_{k=1}^{m}k.k\]\[=(n -m ) \sum_{k=1}^{m}k + \sum_{k=1}^{m}k^2 \]\[=( n- m) (\frac{ m }{ 2 } ) ( 1 + m ) + \frac{ m }{ 6 } (n + 1 )(2n + 1)\]

Answer 3

because
\[\sum_{k=1}^{m} k = \frac{ m }{ 2 }( m + 1)\]
and\[\sum_{k=1}^{m}k^2 = \frac{ m }{ 6 }( m +1 ) ( 2m +1)\]

Answer 4

srry... the last answer should be...
\[=( n - m) ( \frac{ m }{ 2} ) ( m + 1) + \frac{ m }{ 6 } ( m +1) ( 2m + 1)\]