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$$\sigma_3 = \sum_{k=0}^7 (k+2)^3 $$

in the image I have attached you can see an equation which I can not solve. However one student at my university who has solved the equation has added an 8 (+8) to the equation in the next step and has increased the "0" under the summation sign to a 1. I think if you could explain why he/she did that it would be a great help. Thanks.

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  • Can you show us what you've tried? We're more receptive to questions if we see that the poster has put some effort into it. Also, if you didn't mind formatting your question so that we might see the original task, that would be helpful in the future. – scoopfaze Oct 22 '19 at 16:01
  • To answer your second question: he changed the index from $0$ to $1$ because when he plugged in $0$ into $(k+2)^3$, he noticed that $2^3=8$, so now he just needs to add from $k=1$ to $k=7$ and just add $8$. – scoopfaze Oct 22 '19 at 16:03
  • copy paste this sum([(k + 2)^3 for k in range(8)]) into this cite https://sagecell.sagemath.org/ and press "Evaluate". – WhatsUp Oct 22 '19 at 16:42

1 Answers1

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$\sum_{k=0}^{7}(k+2)^3=\sum_{k=0}^{7}(k^3+6k^2+12k+8)= \sum_{k=0}^{7} k^3+6\sum_{k=0}^{7}k^2+12\sum_{k=0}^{7}k+\sum_{k=0}^{7}8$.

$k=0$ doesn't contribute in the first 3 terms, but you must count it in the final as it doesn't depend on $k$.

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