# Summing Series (Sigma Notation)

• I-need-help
The best that you can do is complete the square on it.In summary, the conversation discusses how to subtract and factorize large fractions using the summation series formulae. The interlocutors suggest finding a common denominator and rewriting the fractions before subtracting. They also mention completing the square to factorize a quadratic factor that cannot be factored further with integer coefficients.
I-need-help

a) nƩr2(r-1)r=1

## Homework Equations

Using the summation series formulae...

## The Attempt at a Solution

So far I have got:

r2(r-1) = r3-r2

Ʃr3 = $\frac{1}{4}$n2 (n+1)2

Ʃr2 = $\frac{1}{6}$n(n+1)(2n+1)

Therefore,

Ʃr3-r2 = $\frac{1}{4}$n2 (n+1)2 - $\frac{1}{6}$n(n+1)(2n+1)

But how do I subtract these two massive things from each other??

Thanks.

They aren't that massive. Rewrite them as fractions:
$$\frac{n^2(n+1)^2}{4}-\frac{n(n+1)(2n+1)}{6}$$
Find a common denominator, rewrite both fractions so that both have that common denominator, and then subtract. It will be possible to factor things out.

It's pretty much just subtracting fractions. Get a common denominator:
The least common denominator of 4 and 6 is 12. To change the first denominator to 12, multiply both numerator and denominator by 3.
$$\frac{n^2(n+1)^2}{4}= \frac{3n^2(n+1)^2}{12}$$

To change the second denominator to 12, multiply both numerator and denominator by 4.
$$\frac{4n(n+1)(2n+1)}{12}$$

To subtract now, subtract the numerators:
$$\frac{3n^2(n+1)^2}{12}- \frac{4n(n+1)(2n+1)}{12}$$

You will need to multiply those:
$3n^2(n+1)^2= 3n^2(n^2+ 2n+ 1)= 3n^4+ 6n^3+ 3n^2$
$4n(n+1)(2n+1)= 4n(2n^2+ 3n+ 1)= 8n^3+ 12n^2+ 4n$

and now subtract.

eumyang said:
They aren't that massive. Rewrite them as fractions:
$$\frac{n^2(n+1)^2}{4}-\frac{n(n+1)(2n+1)}{6}$$
Find a common denominator, rewrite both fractions so that both have that common denominator, and then subtract. It will be possible to factor things out.

Yeah ok, I've done that, but the textbook answers want it to be "factorised", and I'm unsure how to do that after subtracting... Thanks for your help.

@HallsOfIvy

Hmm thanks, for the help, I'll try to factorise those now, since the book wants it to be factorised.

You will have a quadratic factor that cannot be factored further (with integer coefficients).

## 1. What is "Sigma Notation"?

Sigma Notation, also known as summation notation, is a mathematical notation used to represent the addition of a series of terms. It uses the Greek letter sigma (Σ) to indicate the sum, and a variable or expression below the sigma to represent the terms being added.

## 2. How is "Sigma Notation" used to sum a series?

To use Sigma Notation to sum a series, you first write out the series in terms of the variable or expression below the sigma. Then, you replace the variable or expression with a starting value and an ending value, and the sigma indicates that you should add all the terms together. For example, the series 1 + 2 + 3 + 4 + 5 can be written as Σn from 1 to 5, where n is the variable representing each term.

## 3. What is the difference between upper and lower limits in "Sigma Notation"?

The upper limit in Sigma Notation represents the last term in the series, while the lower limit represents the first term. For example, Σn from 1 to 5 means that you add all the terms from n=1 to n=5, so the upper limit is 5 and the lower limit is 1.

## 4. Can "Sigma Notation" be used for infinite series?

Yes, Sigma Notation can be used for both finite and infinite series. However, for infinite series, the notation is typically written as Σn from 1 to ∞, where ∞ represents infinity as the upper limit.

## 5. What are some common properties of "Sigma Notation"?

Some common properties of Sigma Notation include the ability to manipulate the limits, the ability to factor out constants, and the ability to add or subtract series with the same limits. It also follows the commutative and associative properties of addition, meaning that changing the order or grouping of terms will not affect the sum.

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