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.

I have the following summation: $$\sum_{i=0}^n\left(6in + n(8n+2)\right).$$ Evaluating I get \begin{align*} 6n\left(\frac{n(n+1)}{2}\right) + n(8n+2) = 3n(n(n+1)) + n(8n+2)\\ &=3n^2(n+1) + n(8n+2)\\ &=3n^3 + 3n^2 + 8n^2 + 2n\\ &=3n^3 + 11n^2 + 2n. \end{align*}

Is this correct?

share|improve this question
1  
Are you starting with $$\left(\sum_{i=0}^{n}6in\right)+n(8n+2)$$ or $$\sum_{i=0}^{n}\left(6in+n(8n+2)\right)?$$ –  Isaac Mar 3 '11 at 22:04
    
the second one. –  Krysten Mar 3 '11 at 22:05
    
I know that what you want is the value of the sum, but isn't $\sum_{i=0}^n\left(6in + n(8n+2)\right)$ a closed form? –  TCM Mar 4 '11 at 1:44

1 Answer 1

up vote 3 down vote accepted

Your first step is wrong.

$$\sum_{i=0}^{n} (6in + n(8n+2)) = \sum_{i=0}^{n} (6in) + \sum_{i=0}^{n} n(8n+2)$$

You wrote

$$\sum_{i=0}^{n} (6in) = 6n \sum_{i=0}^{n} i = 6n^2(n+1)/2$$

but missed this one:

$$\sum_{i=0}^{n} n(8n+2) = n(8n+2) \sum_{i=0}^n 1 = n(n+1)(8n+2)$$

and wrote

$$n(8n+2)$$

instead.

share|improve this answer
    
in the second summation, are you adding n+1 because the index starts at 0 and not 1? –  Krysten Mar 3 '11 at 22:09
    
@kry: Yes, exactly! You are adding n+1 times. –  Aryabhata Mar 3 '11 at 22:10
    
but when evaluating the first summation, it does not need to be multiplied by (n+1) because you are plugging in for i directly? –  Krysten Mar 3 '11 at 22:15
    
@Kry: I have edited the answer to clarify it a bit. Since $n$ is independent of $i$, you can take it out of the $\sum$. –  Aryabhata Mar 3 '11 at 22:19
    
thanks! that cleared it up –  Krysten Mar 3 '11 at 22:21

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.