Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 -

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

5 Answers 5

up vote 4 down vote accepted

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 $$S_n = (-1)^{n+1}(2n+1) + 1$$

share|improve this answer

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

share|improve this answer
2  
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 added some more information at my answer.. –  Mary Star Sep 1 at 12:13
    
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 $$(-1)^{2k}=1,\ \ (-1)^{2k+1}=-1.$$

share|improve this answer

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.

share|improve this answer

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

enter image description here

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.