Summation Notation Formulas

**UnstoppableBeast** · August 19th 2011, 05:00 PM

I'm sorry that I don't know the code for sigma notation for the forums but my question is a fairly straight forward one.

I've been give

$End = 35, Start = 1 (3n^2+5n-2)$

$End = 50, Start = 1 (n-4)^3$

I would be a serious pain in the butt to input each number into the equation and add them all up, so I'd like to know if anyone knew the formulas? I have found the formulas.

At this url, and my question is...accoding to the formulas, I start with a negative number? Considering the number without an exponent or variable (-2) is negative.

**SammyS** · August 19th 2011, 05:17 PM

Originally Posted by UnstoppableBeast

I'm sorry that I don't know the code for sigma notation for the forums but my question is a fairly straight forward one.

I've been given

$End = 35, Start = 1 (3n^2+5n-2)$

$End = 50, Start = 1 (n-4)^3$

I would be a serious pain in the butt to input each number into the equation and add them all up, so I'd like to know if anyone knew the formulas? I have found the formulas.

At this url, and my question is...accoding to the formulas, I start with a negative number? Considering the number without an exponent or variable (-2) is negative.

Use "\sum_{n=1}^{35}(3n^2+5n-2)" along with TEX tags to get:

$\sum_{n=1}^{35}(3n^2+5n-2)$

**Prove It** · August 19th 2011, 06:19 PM

Originally Posted by UnstoppableBeast

I'm sorry that I don't know the code for sigma notation for the forums but my question is a fairly straight forward one.

I've been give

$End = 35, Start = 1 (3n^2+5n-2)$

$End = 50, Start = 1 (n-4)^3$

I would be a serious pain in the butt to input each number into the equation and add them all up, so I'd like to know if anyone knew the formulas? I have found the formulas.

At this url, and my question is...accoding to the formulas, I start with a negative number? Considering the number without an exponent or variable (-2) is negative.

1. $\displaystyle \sum_{n = 1}^{35}{\left(3n^2 + 5n - 2\right)} = 3\sum_{n = 1}^{35}{\left(n^2\right)} + 5\sum_{n = 1}^{35}{(n)} - 2\sum_{n = 1}^{35}{(1)}$ .

There are formulas to evaluate the first two summands, and the third is simply $\displaystyle 35 \times 1$ .

2.
$\displaystyle \begin{align*} \sum_{n = 1}^{50}{\left(n - 4\right)^3} &= (-3)^3 + (-2)^3 + (-1)^3 + 0^3 + 1^3 + 2^3 + \dots + 56^3 \\ &= -27 - 8 - 1 + 0 + \sum_{n = 1}^{56}{n^3} \\ &= -36 + \sum_{n = 1}^{56}{n^3} \end{align*}$

There is a formula to sum cubes as well

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