Sum of a series of numbers

**lynch-mob** · June 3rd 2008, 02:38 AM

Calculate the sum of the squares of following numbers:

2,5,8,11,...,(3n-4),(3n-1)

???

**wingless** · June 3rd 2008, 03:08 AM

The sequence is $a_k = 3k - 1$

So you are trying to find $\sum_{k=1}^{n}(3k-1)^2$

$\sum_{k=1}^{n}(9k^2 - 6k +1)$

Now calculate $\sum_{k=1}^{n}9k^2 - \sum_{k=1}^{n}6k + \sum_{k=1}^{n}1$ .

**lynch-mob** · June 3rd 2008, 04:11 AM

Can´t believe, such a fast reply.

Thanks for the formula. My problem now is how to calculate the sum. I mean the n is not specified so the sum will be humongous. Please help me with the rest?

**Soroban** · June 3rd 2008, 04:25 AM

Hello, lynch-mob!

If you're familiar with summation formulas,
. . wingless has an excellent solution.

Here's a very primitive method for constructing the formula.

Calculate the sum of the squares of following numbers:
. . $2, 5, 8, 11, \hdots, (3n-1)$

Consider the partial sums of: . $f(n) \;=\;2^2 + 5^2 + 8^2 + 11^2 + \hdots$

We have the sequence: . $4,\: 29,\: 93,\: 214,\: 410,\: 699,\: 1099,\: \hdots$

Take the difference of consecutives terms,
. . take the differences of the differences, and so on . . .

$\begin{array}{ccccccccccccccc}\text{Sequence: } & 4 && 29 && 93 && 214 && 410 && 699 && 1099 \\ \text{1st diff.} & & 25 && 64 && 121 && 196 && 289 && 400 \\ \text{2nd diff} & & & 39 && 57 && 75 && 93 && 111 \\ \text{3rd diff} & & & & 18 && 18 && 18 && 18\end{array}$

We find that the third differences are constant.
. . Hence, the generating function is a cubic.

The general cubic function is: . $f(n) \;=\;an^3 + bn^2 + cn + d$

Use the first four known values to form a system of equations.

$\begin{array}{ccccc} f(1) = 4: & a + b + c + d &=& 4 & {\color{blue}[1]} \\ f(2) = 29: & 8a + 4b + 2c + d &=& 29 & {\color{blue}[2]} \\ f(3) = 93: & 27a + 9b + 3c + d &=& 93 & {\color{blue}[3]} \\ f(4) = 214: & 64a + 16b + 4c + d &=& 214 & {\color{blue}[4]} \end{array}$

$\begin{array}{ccccc}\text{Subtract [2] - [1]:} & 7a + 3b + c &=& 25 & {\color{blue}[5]} \\ \text{Subtract [3] - [2]:} & 19a + 5b + c &=& 64 & {\color{blue}[6]} \\ \text{Subtract [4] - [3]:} & 37a + 7b + c &=& 121 & {\color{blue}[7]} \end{array}$

$\begin{array}{ccccc}\text{Subtract [6] - [5]:} & 12a + 2b &=& 39 & {\color{blue}[8]} \\ \text{Subtract [7] - [6]:} & 18a + 2b &=& 57 & {\color{blue}[9]} \end{array}$

$\text{Subtract [9] - [8]: }\;\;6a \:=\:18\quad\Rightarrow\quad\boxed{ a \:=\:3}$

Substitute into [8]: . $36 + 2b \:=\:39\quad\Rightarrow\quad\boxed{ b \:=\:\frac{3}{2}}$

Substitute into [5]: . $21 + \frac{9}{2} + c \:=\:25\quad\Rightarrow\quad\boxed{ c \:=\:-\frac{1}{2}}$

Substitute into [1]: . $3 +\frac{3}{2} - \frac{1}{2} + d \:=\:4\quad\Rightarrow\quad\boxed{d \:=\:0}$

Hence: . $f(n) \;=\;3n^3 + \frac{3}{2}n^2 - \frac{1}{2}n \quad\Rightarrow\quad\boxed{ f(n) \;=\;\frac{n}{2}(6n^2 + 3n-1)}\quad\hdots$ There!

**lynch-mob** · June 3rd 2008, 04:53 AM

Umm...when i put number 4 in the function i get 214. This is the correct answer for the sum of that sequence. So the correct answer when n is unspecified is the actual formula?

I have trouble understanding wingless summation formula, i hope somebody could explain that one for me.

I find it so hard to understand that there is a forum like this where answers (math answers) are given the same day as the question. How do you write math functions and symbols here. It´s so much easier to understand than text like this 2*14(5X-3)/4*^2.

**lynch-mob** · June 3rd 2008, 05:49 AM

So now I understand. Your summation formula gives me the number I get when I square the n:th number. Lets say the third number in the sequence is 8.

$9*(3)^2-(6*3)+1 = 64$

... which is number 8 squared. But how do I sum all the previous and consecutive squared numbers up to the n:th number.

Thanks!

**Aryth** · June 3rd 2008, 06:02 AM

You just use the rule in summation notation:

$\sum_{k=1}^n f(k) = \sum_{k=1}^n \frac{k}{2}(6k^2 + 3k - 1)$

**wingless** · June 3rd 2008, 06:44 AM

The letter $\sum$ (sigma) means sum.

$\sum_{k=n}^m f(k) = f(n) + f(n+1) + f(n+2) + ... + f(m-1) + f(m)$

The sum has m-n+1 terms.

For example, $\sum_{k=0}^{5}(2k+1) = 1 + 3 + 5 + 7 + 9 + 11 = 36$

A few identities and formulas with sigma:

$\sum_{k=n}^m a\cdot f(k) = a\cdot \sum_{k=n}^m f(k)$

$\sum_{k=1}^n k = \frac{n(n+1)}{2}$

$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$

$\sum_{k=1}^n k^3 = \left ( \frac{n(n+1)}{2} \right )^2$

Using these formulas, try to find
$\sum_{k=1}^{n}9k^2 - \sum_{k=1}^{n}6k + \sum_{k=1}^{n}1$

**Soroban** · June 3rd 2008, 08:03 AM

Hello, lynch-mob!

Originally Posted by wingless

The sequence is $a_k = 3k - 1$
So you are trying to find $\sum_{k=1}^{n}(3k-1)^2$
$\sum_{k=1}^{n}(9k^2 - 6k +1)$
Now calculate $\sum_{k=1}^{n}9k^2 - \sum_{k=1}^{n}6k + \sum_{k=1}^{n}1$

We have: . $S \;=\;9\sum^n_{k=1}k^2 - 6\sum^n_{k=1}k + \sum^n_{k=1}1\;\;{\color{blue}[1]}$

These are standard summation formulas:

. . $\boxed{\sum^n_{k=1}k^2 \;=\;\frac{n(n+1)(2n+1)}{2}} \qquad \boxed{\sum^n_{k=1} k \;=\;\frac{n(n+1)}{2}}\qquad \boxed{\sum^n_{k=1}1 \;=\; n}$

Substitute them into [1]: . $S \;\;=\;\;9\!\cdot\!\frac{n(n+1)(2n+1)}{6} \:- \:6\!\cdot\!\frac{n(n+1)}{2} \:+ \:n$

. . and we get: . $S \;\;=\;\;\frac{n}{2}(6n^2 + 3n-1)$

**lynch-mob** · June 3rd 2008, 11:35 PM

Thank you so much guys. Now I get it. All the time I was wondering how can I calculate the formula when n is undefined. So the answer will be a function where n is the n:th number in the series and S is the total sum of all the squared numbers.

This must be simple to you guys, but such a crash course in sums to me. I think I´m finally starting to get it!

Thanks wingless, soroban and aryth!

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Sum of a series of numbers

Thanks man

Thanks guys

Aaaah wingless

Aaaah

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