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I am learning how to evaluate summation and got stuck on evaluating this summation: It would be great if you could help me through this.

$$ \sum_{i=0}^{n} \sum_{i=0}^{m}3^{i+j} $$

also after a bit of scratch work, I have found that this summation will be usefull: $$ 1+r+r^2+r^3+r^4+...+r^n = \frac{1-r^{n+1}}{1-r} $$

Again, any lead into this will help me out a lot! Thanks!

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  • $\begingroup$ Do you have even tried ? This is simply $\sum_{i=0}^n3^i\sum_{j=0}^m3^j$ $\endgroup$ – mwoua Mar 30 '13 at 9:01
  • $\begingroup$ @mwoua ahh dang that's it! Just need to separate them. Sorry im a newbie to summations! $\endgroup$ – Krimson Mar 30 '13 at 18:13
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$$ \sum_{i=0}^n \sum_{j=0}^m3^{i+j}$$ $$=\sum_{i=0}^n 3^i \left(\sum_{j=0}^m 3^j\right)\text{ as $i$ is independent of $j$}$$

$$=\sum_{i=0}^n 3^i\left(\frac{3^{m+1}-1}{3-1}\right)$$

$$=\left(\frac{3^{m+1}-1}2\right)\sum_{i=0}^n 3^i\text{ as $m$ is independent of $i,n$}$$

$$=\left(\frac{3^{m+1}-1}2\right)\cdot\left( \frac{3^{n+1}-1}2\right)$$

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Just apply your formula. For a nested sum, you can focus on one summation index while keeping the other fixed. In this case, write the sum as

$$\sum_{i=0}^n 3^i \sum_{j=0}^m 3^j$$

So you just have 2 separate sums of identical form to evaluate; the answer you seek is the product of these sums.

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