Joel French

2022-07-01

How to find the values of these kind of summations:
$\sum _{i=0}^{6}\left(6-i\right)\phantom{\rule{thickmathspace}{0ex}}\ast \phantom{\rule{thickmathspace}{0ex}}\sum _{j=1}^{6}\left(7-j\right)\phantom{\rule{thickmathspace}{0ex}}\ast \phantom{\rule{thickmathspace}{0ex}}\sum _{k=2}^{7}\left(8-k\right)\phantom{\rule{thickmathspace}{0ex}}\ast \phantom{\rule{thickmathspace}{0ex}}\sum _{\ell =3}^{8}\left(9-\ell \right)$

Jaruckigh

Use that
$\begin{array}{rl}\sum _{t=a}^{b}\left(c-t\right)& =\left(\sum _{t=a}^{b}c\right)-\left(\sum _{t=a}^{b}t\right)\\ \\ & =\left(\sum _{t=a}^{b}c\right)-\left(\sum _{s=0}^{b-a}\left(s+a\right)\right)\\ \\ & =\left(\sum _{t=a}^{b}c\right)-\left(\sum _{s=0}^{b-a}s\right)-\left(\sum _{s=0}^{b-a}a\right)\\ \\ & =\left(b-a+1\right)c-\left(b-a+1\right)a-\left(\sum _{s=0}^{b-a}s\right)\\ \\ & =\left(b-a+1\right)\left(c-a\right)-\left(\sum _{s=0}^{b-a}s\right)\\ \\ & =\left(b-a+1\right)\left(c-a\right)-\frac{\left(b-a\right)\left(b-a+1\right)}{2}\\ \\ & =\left(b-a+1\right)\left(c-a-\frac{b}{2}+\frac{a}{2}\right)\\ \\ & =\left(b-a+1\right)\left(c-\frac{b}{2}-\frac{a}{2}\right)\end{array}$
for each term. For example,
$\sum _{i=0}^{6}\left(6-i\right)=6+5+4+3+2+1=\overline{)\text{21}}=7\cdot 3=\left(6-0+1\right)\left(6-\frac{6}{2}-\frac{0}{2}\right)\phantom{\rule{2em}{0ex}}✓$

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