# How to calculate these summations?

How to find the values of these kind of summations:

$$\large\sum_{i=0}^6(6-i)\;\ast\;\sum_{j=1}^6(7-j)\;\ast\;\sum_{k=2}^7(8-k)\;\ast\;\sum_{\ell=3}^8(9-\ell)$$

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@m.woj * means multiplication here. – Eight Jun 8 '12 at 6:49
abhinav8, Zev Thanks. (Question was: what does asterisk mean in that context?) – data Jun 8 '12 at 6:51

Use that \begin{align}\sum_{t=a}^b(c-t)&=\left(\sum_{t=a}^bc\right)-\left(\sum_{t=a}^b t\right)\\\\&=\left(\sum_{t=a}^bc\right)-\left(\sum_{s=0}^{b-a} (s+a)\right)\\\\ &=\left(\sum_{t=a}^bc\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-\tfrac{b}{2}+\tfrac{a}{2})\\\\&=(b-a+1)(c-\tfrac{b}{2}-\tfrac{a}{2})\end{align} for each term. For example, $$\sum_{i=0}^6(6-i)=6+5+4+3+2+1=\fbox{21}=7\cdot 3=(6-0+1)(6-\tfrac{6}{2}-\tfrac{0}{2})\qquad\checkmark$$

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you seems to be wrong in last two steps of your equation, (c-a-(b/2)+(a/2)) should be equal to (c-(b/2)-(a/2)) but you have got (c-(b/2)+(a/2)).pls check and correct me if i am wrong. – Eight Jun 8 '12 at 6:43
@abhinav8: Thanks for catching that! You're correct, I made a careless error. I've edited my answer with the correct expression. – Zev Chonoles Jun 8 '12 at 6:45

Or the formula:

$$\sum_{i=k_1}^{k_2} (p-i) = (k_2-k_1+1)(p-\frac{1}{2}(k_1+k_2)).$$

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The answer to the original question is 194,481. – copper.hat Jun 8 '12 at 6:29

Hint: Use the following formula.

• $\displaystyle \sum\limits_{i=1}^{n} i = \frac{n \cdot (n+1)}{2}$.

• $\displaystyle \sum\limits_{i=0}^{6} (6-i) = \sum\limits_{i=0}^{6} (6 - i) = 6 \cdot \bigl(1+1+1+1+1+1+1\bigr)- \sum_{i=0}^{6} i =6 \times 7 - \frac{6 \cdot 7}{2}$

Similarly you can calculate the other summation's as well and then multiply them.

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Huh? How would that work in this case? – Gigili Jun 8 '12 at 6:09
@Gigili: I have edited my answer. – user9413 Jun 8 '12 at 6:12
Should the answer be $1+...+6 = 21$? – copper.hat Jun 8 '12 at 6:18
@copper.hat: Yeah. $21 = \frac{6 \cdot 7}{2}$ – user9413 Jun 8 '12 at 6:19
I may be completely mistaken, but you value of $\sum_{i=0}^6 (6-i)$ seems completely wrong. Why would $i$ start first at 0 then 1? Why would $\sum_{i=1}^6 6 = 6 \cdot (1+2+3+4+5+6)$? – Najib Idrissi Jun 8 '12 at 6:28