Joel French

2022-07-01

How to find the values of these kind of summations:

$\sum _{i=0}^{6}(6-i)\phantom{\rule{thickmathspace}{0ex}}\ast \phantom{\rule{thickmathspace}{0ex}}\sum _{j=1}^{6}(7-j)\phantom{\rule{thickmathspace}{0ex}}\ast \phantom{\rule{thickmathspace}{0ex}}\sum _{k=2}^{7}(8-k)\phantom{\rule{thickmathspace}{0ex}}\ast \phantom{\rule{thickmathspace}{0ex}}\sum _{\ell =3}^{8}(9-\ell )$

$\sum _{i=0}^{6}(6-i)\phantom{\rule{thickmathspace}{0ex}}\ast \phantom{\rule{thickmathspace}{0ex}}\sum _{j=1}^{6}(7-j)\phantom{\rule{thickmathspace}{0ex}}\ast \phantom{\rule{thickmathspace}{0ex}}\sum _{k=2}^{7}(8-k)\phantom{\rule{thickmathspace}{0ex}}\ast \phantom{\rule{thickmathspace}{0ex}}\sum _{\ell =3}^{8}(9-\ell )$

Jaruckigh

Beginner2022-07-02Added 11 answers

Use that

$\begin{array}{rl}\sum _{t=a}^{b}(c-t)& =\left(\sum _{t=a}^{b}c\right)-\left(\sum _{t=a}^{b}t\right)\\ \\ & =\left(\sum _{t=a}^{b}c\right)-(\sum _{s=0}^{b-a}(s+a))\\ \\ & =\left(\sum _{t=a}^{b}c\right)-\left(\sum _{s=0}^{b-a}s\right)-\left(\sum _{s=0}^{b-a}a\right)\\ \\ & =(b-a+1)c-(b-a+1)a-\left(\sum _{s=0}^{b-a}s\right)\\ \\ & =(b-a+1)(c-a)-\left(\sum _{s=0}^{b-a}s\right)\\ \\ & =(b-a+1)(c-a)-\frac{(b-a)(b-a+1)}{2}\\ \\ & =(b-a+1)(c-a-{\textstyle \frac{b}{2}}+{\textstyle \frac{a}{2}})\\ \\ & =(b-a+1)(c-{\textstyle \frac{b}{2}}-{\textstyle \frac{a}{2}})\end{array}$

for each term. For example,

$\sum _{i=0}^{6}(6-i)=6+5+4+3+2+1=\overline{)\text{21}}=7\cdot 3=(6-0+1)(6-{\textstyle \frac{6}{2}}-{\textstyle \frac{0}{2}})\phantom{\rule{2em}{0ex}}\u2713$

$\begin{array}{rl}\sum _{t=a}^{b}(c-t)& =\left(\sum _{t=a}^{b}c\right)-\left(\sum _{t=a}^{b}t\right)\\ \\ & =\left(\sum _{t=a}^{b}c\right)-(\sum _{s=0}^{b-a}(s+a))\\ \\ & =\left(\sum _{t=a}^{b}c\right)-\left(\sum _{s=0}^{b-a}s\right)-\left(\sum _{s=0}^{b-a}a\right)\\ \\ & =(b-a+1)c-(b-a+1)a-\left(\sum _{s=0}^{b-a}s\right)\\ \\ & =(b-a+1)(c-a)-\left(\sum _{s=0}^{b-a}s\right)\\ \\ & =(b-a+1)(c-a)-\frac{(b-a)(b-a+1)}{2}\\ \\ & =(b-a+1)(c-a-{\textstyle \frac{b}{2}}+{\textstyle \frac{a}{2}})\\ \\ & =(b-a+1)(c-{\textstyle \frac{b}{2}}-{\textstyle \frac{a}{2}})\end{array}$

for each term. For example,

$\sum _{i=0}^{6}(6-i)=6+5+4+3+2+1=\overline{)\text{21}}=7\cdot 3=(6-0+1)(6-{\textstyle \frac{6}{2}}-{\textstyle \frac{0}{2}})\phantom{\rule{2em}{0ex}}\u2713$

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