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According to this link $$ \sum\limits_{k=j}^{i+j} (j+i-k) $$ is $$ \sum\limits_{k=1}^i (k) $$ Can someone write the Sum formula(s) which was used in this transformation?

pierre
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2 Answers2

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Writing it, \begin{align*} &\;\;\; \sum\limits_{k=j}^{i+j} (j+i-k) \\ &= ((j + i - j) + (j + i - j - 1) + \dots + 1 + 0)\\ &= (i + (i-1) + ... + 1) \\ &= \sum_{k=1}^i k \end{align*}

Shifting index $k \to k + j$, $$\sum\limits_{k=j}^{i+j} (j+i-k)= \sum_{k=0}^i (j + i - (k+j)) = \sum_{k=0}^i (i-k) = \sum_{k=1}^i k$$

S L
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  • Almost there but I don't understand according which rule you have transformed $$ \sum\limits_{k=0}^i (i-k) $$ to $$ \sum\limits_{k=1}^i (k) $$ – pierre Sep 25 '13 at 20:18
  • @pierre either you count it from back or front, it's just same. the sum is like that. it is just adding the sum from back instead of front. – S L Sep 25 '13 at 20:19
  • @pierre or just try this $$\sum_{k=1}^i i - \sum_{k=1}^i k = i(i+1) - \frac{i(i+1)}{2} = \frac{i(i+1)}{2}$$ – S L Sep 25 '13 at 20:21
  • Ok, one more think, why $$ \sum\limits_{k=1}^i i = i(i + 1) $$

    I think i in this case is constant and according to next rule: $$ \sum\limits_{k=1}^i c = ci $$

    it should be i^2

     – pierre Sep 26 '13 at 15:01
  • @pierre yes you are absolutely correct ... but my index begins from 0. $\displaystyle \sum_{k=0}^i c = c(i+1)$ Woops!! sorry ... my mistake. on earlier sum ... the index begins from zero. – S L Sep 26 '13 at 15:02
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Let's do things a bit differently; maybe that will make it clearer.

All that happened here is re-indexing of the terms of the sequence. In other words, all of the same terms appear... we're just assigning them a different index number, and possibly re-ordering them.

You started off here with $$ \sum_{k=j}^{i+j}(j+i-k). $$ Notice that $j+i$ never changes. Let's define a new index of summation, say $t$, by $t=j+i-k$.

Notice that when we plug in $k=j$, we get $t=i$; when we plug in $k=j+1$, we get $t=i-1$; and so on, and so forth, until we get to $k=j+i$, in which case $t=0$.

So, summing over all values of $k$ between $j$ and $j+i$ is the same as summing over all values of $t$ between $0$ and $i$. All that remains is to rewrite our index $k$ term of the summand in terms of $t$; since $t=j+i-k$, this term is precisely $t$. So, we conclude that $$ \sum_{k=j}^{j+i}(j+i-k)=\sum_{t=0}^{i}t. $$

Does that help?

Nick Peterson
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