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Why is this equivalence true?

$$\sum_{i=0}^{n-5} 4(n-i-5)^3 = 4 \sum_{i=5}^n (n-i)^3$$ With the first sum I would make an index shift by 5 and would get: $$4 \sum_{i=5}^n (n-i-10)^3.$$ My questions are thus: Why is the first equivalence true? What is the procedure and why is my index shift wrong?

dfeuer
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Elternhaus
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  • Put $i=0$ in the term of the first sum. You get $4(n-5)^3$. Put $i=5$ in the term of your erroneous sum at the end. You get $4(n-15)^3$. – OR. Dec 08 '13 at 18:39
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    You tried to do to much in one step. When you do your index shift, put the modified $i$ in parentheses with the appropriate shift. Then simplify. Don't try to do it all in your head. There are people who can do huge calculations in their heads. For the rest of us, there's paper. – dfeuer Dec 08 '13 at 18:42
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    Shifting the sum is just change of variable. Put $i-5$ everywhere (!!!) instead of $i$. If we do just that, we get $\sum_{i-5=0}^{i-5=n-5}4(n-(i-5)-5)^3$, which is equal to $\sum_{i=5}^{i=n}4(n-i+5-5)^3$. – OR. Dec 08 '13 at 18:44

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$$\sum_{i=0}^{n-5}4\,(n-i-5)^3=4\cdot\sum_{i=0}^{n-5}(n-i-5)^3=4\cdot\sum_{i=0}^{n-5}[n-(i+5)]^3=4\cdot\sum_{j=5}^n(n-j)^3$$

Lucian
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