# How to describe 4-8+12-16+20-24+28 using summation ($\Sigma$) notation?

Can anyone find the ∑ summation for this please?

4-8+12-16+20-24+28

It seems to be going up by steps of 4, but I can't seem to get how I should write it down, since it uses both + and -

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$\sum_{k=1}^7 (-1)^{k-1}(4k)$ –  Nilan Sep 1 at 11:39

The partial sums$$S_n = \sum_{r=1}^n 4r(-1)^{r+1}$$ are $4, -4, 8, -8, 12, -12,\ldots$. Subtract $1$ to get $3,-5,7,-9,11,-13,\ldots$ which is $(-1)^{n+1}(2n+1)$. Thus $$S_n = (-1)^{n+1}(2n+1) + 1$$

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$$4-8+12-16+20-24+28=4(1-2+3-4+5-6+7)=4((-1)^{1+1}\cdot 1+(-1)^{2+1}\cdot 2+(-1)^{3+1}\cdot 3+(-1)^{4+1} \cdot 4+(-1)^{5+1} \cdot 5+(-1)^{6+1} \cdot 6+(-1)^{7+1} \cdot 7)=4 \sum_{i=1}^{7}(-1)^{i+1}i=4 \left ( \sum_{i=0}^{3} (2i+1)-\sum_{i=1}^{3} 2i\right )=4 \left( 2 \sum_{i=0}^{3} i +\sum_{i=0}^{3} 1-2\sum_{i=1}^{3} i\right )=4 \left( 2 \sum_{i=1}^{3} i +\sum_{i=0}^{3} 1-2\sum_{i=1}^{3} i\right ) \\ =4 \sum_{i=0}^{3} 1=4 \cdot (3+1)=4 \cdot 4=16$$

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Uh...what? Do you think $4\cdot 7$ is $4\cdot (-7)^8$? You've got typos here. Maybe you mean $4\cdot 7\cdot(-1)^8$ instead? I find it incredible that this answer received 4 upvotes. –  MPW Sep 1 at 11:41
Ohh...you're right! I edited my answer.. –  Mary Star Sep 1 at 11:43
How does this help? –  TonyK Sep 1 at 11:46
I rather think the OP is asking for a general expression in terms of $n$. (And even if I'm wrong, there are easier ways to evaluate $4-8+12-16+20-24+28$ !) –  TonyK Sep 1 at 13:03

HINT : For $k\in\mathbb N$, we have $$(-1)^{2k}=1,\ \ (-1)^{2k+1}=-1.$$

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Let $S_n=\sum_{k=1}^n 4(-1)^{k+1}k$, for $n\geq1$.

Then

$$S_{2n}-S_{2(n-1)}=4(-1)^{2n+1}2n+4(-1)^{2n}(2n-1)=-8n+8n-4=-4$$

And

$$S_{2n+1}-S_{2n-1}=4(-1)^{2n+2}(2n+1)+4(-1)^{2n+1}(2n)=8n+4-8n=4$$

Hence, for $n\geq 1$,

$$S_{2n}=S_2+\sum_{k=2}^n S_{2k}-S_{2k-2}=-4-4(n-1)=-4n$$

And

$$S_{2n-1}=S_1+\sum_{k=1}^{n-1} S_{2k+1}-S_{2k-1}=4+4(n-1)=4n$$

All in all,

$$S_n=4(-1)^{n+1}\left\lceil\frac{n}{2}\right\rceil$$

Where $\lceil x\rceil$ denotes the ceiling function.

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This may be redundant coming after TonyK's very nice answer above and answers by others, but here it is anyway.

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