Calculating the sum of a series of cubes

[John3509]

Mentor note: Moved from technical math section, so missing the homework template

filling in the boxes is easy, 1,3,6,10,15
second is just squares of that,
1,9,36,100,225

But how I anyone supposed to find and expression for this?
This is from a textbook on elementary algebra, the specific chapter is for solving equations where the variable appears on both sides....
I don't see how you could find an expression with any of the material that has been covered thus far.

I did manage to find an equation for table1, (.5)x²+(.5)x using some trick from kinematics
but it will not work for table2.

So how are you supposed to find an expression for table 1 and 2?

 

[Mark44]

Notice that the sums for the second table can be obtained by adding the numbers that are being cubed, and then squaring that number. IOW, for the sum of the cubes of the first k integers, the sum will be $(\sum_{n=1}^k n)^2$. Can you work with that to come up with a more simplified expression?

 

[John3509]

Notice that the sums for the second table can be obtained by adding the numbers that are being cubed, and then squaring that number. IOW, for the sum of the cubes of the first k integers, the sum will be $(\sum_{n=1}^k n)^2$. Can you work with that to come up with a more simplified expression?

Not really. ((.5)x²+(.5)x)² Is all I got. But that requires having that equation already, which I still don't get how the book expects you to figure that out, because the way I did certainly isn't it. And while I know what summation notation it is hasn't been covered in the book yet eighter.

 

[Mark44]

$((.5)x^2+(.5)x)^2$ is the same as $(\frac{x(x + 1)}2)^2$. Does your book show you how to get a formula for $\sum_{k= 1}^n k$? IOW, for $1 + 2 + 3 + \dots + n$? That sum is $\frac{n(n + 1)}2$.

To get a formula for the sum of the cubes of the first n integers, look at the pattern you've developed for the results of Table 2.
$1^3 = 1$
$1^3 + 2^3 = 9$
$1^3 + 2^3 + 3^3 = 36$
and so on.
The pattern seems to be "add the numbers being cubed, and then square that result." Does this pattern hold for longer sums?

 

[mcastillo356]

Hi
$$S=\displaystyle\sum_{i=1}^{n}\,i=1+2+3+\cdots=\displaystyle\frac{n(n+1)}{2}$$

Trick:

$S=\,1\,+\,2\,\,+3\,+\cdots+\,(n-1)\,+\,n$
$S=\,n\,+\,(n-1)\,+\,(n-2)\,+\cdots+\,2\,+\,1$
________________________________

$$2S=(n+1)+(n+1)+(n+1)+\cdots+(n+1)+(n+1)=n(n+1)\Rightarrow{S=\displaystyle\frac{n(n+1)}{2}}$$

For the second series, I really don't know if it would work: arithmetic laws turn into the following rule of linearity for finite sums. If $A$ and $B$ are constants, then

$$\displaystyle\sum_{i=m}^{n}{\,{(Af(i))+Bg(i))}=A\,\displaystyle\sum_{i=m}^{n}{\,{f{i}}}\,+\,B\,\displaystyle\sum_{i=m}^{n}}{\,{g{i}}}$$

I would try to use @Mark44 and this last rule of linearity.

See you later!

 

[pbuk]

I think everyone is missing what the question is looking for.

You have noticed that the numbers on the RHS of Table 2 are the squares of the numbers on the RHS of Table 1. The answer to question 2 is simply to write this down: $\left ( 1 + 2 \, + \, ... + \, x \right )^2$

 

[OmCheeto]

Not really. ((.5)x²+(.5)x)² Is all I got. But that requires having that equation already, which I still don't get how the book expects you to figure that out, because the way I did certainly isn't it. And while I know what summation notation it is hasn't been covered in the book yet eighter.

I feel your pain

	power + 1	y1	y2	y3	y4	y5	y6
	10
	9
x^7	8						1/7
x^6	7					1/6	1/2
x^5	6				1/5	1/2	1/2
x^4	5			1/4	1/2	5/12
x^3	4		1/3	1/2	1/3		-1/6
x²	3	1/2	1/2	1/4		-1/12
x	2	1/2	1/6		-1/30		1/42

and also your fascination.

So many patterns.

 

[Mark44]

I feel your pain

	power + 1	y1	y2	y3	y4	y5	y6
	10
	9
x^7	8						1/7
x^6	7					1/6	1/2
x^5	6				1/5	1/2	1/2
x^4	5			1/4	1/2	5/12
x^3	4		1/3	1/2	1/3		-1/6
x²	3	1/2	1/2	1/4		-1/12
x	2	1/2	1/6		-1/30		1/42

and also your fascination.

So many patterns.

I don't understand what this table is supposed to represent, nor what it has to do with the question in this thread.

 

[OmCheeto]

I don't understand what this table is supposed to represent, nor what it has to do with the question in this thread.

It's the discrete solutions to tables 1 and 2, plus 4 other tables without using the ∑ function. Unfortunately, like the OP, it just gives me a generalized pattern for higher powered solutions. I had to cheat to get the first few patterns and then use curve fitting to get the last.

Columns y1 and y3 are the solutions to the OP's tables 1 & 2:
for table 1: $y = \frac 1 2 x^2 + \frac 1 2 x$
for table 2: $y = \frac 1 4 x^4 + \frac 1 2 x^3 + \frac 1 4 x^2$

I suppose it might have been clearer had I transposed the axes in my table.

And as far as my cheating goes, I simply plotted the numbers for each table of powers in a spreadsheet and told the spreadsheet to come up with a polynomial solution, which it is able to do for powers up to 6. But by that point a pattern had exposed itself, so getting an equation that fit for my 'y6' which required an 'x^7' just required a bit of curve fitting.

for a given $x^z$ table, $y = \frac 1 z x^{z+1} + \frac 1 2 x^z + \frac {z-1} {12} x^{z-1}$ works for $x^2$ & $x^3$.
This pattern continues for higher powers except that new factors following different patterns are added.
As to how to solve this problem without the non-methods of the OP and myself, I'll sit and wait for more clues.

 

[mcastillo356]

You have noticed that the numbers on the RHS of Table 2 are the squares of the numbers on the RHS of Table 1. The answer to question 2 is simply to write this down: $\left ( 1 + 2 \, + \, ... + \, x \right )^2$

Given a summation, we often wish to replace it with an algebraic equation with the same value as the summation. This is known as a closed-form solution, and the process of replacing the summation with its closed form solution is known as solving the summation.

$$S=\displaystyle\sum_{i=1}^{n}\,i=\displaystyle\frac{n(n+1)}{2}$$

¿Can we elevate this closed-form solution to its second power?

 

[John3509]

Given a summation, we often wish to replace it with an algebraic equation with the same value as the summation. This is known as a closed-form solution, and the process of replacing the summation with its closed form solution is known as solving the summation.

$$S=\displaystyle\sum_{i=1}^{n}\,i=\displaystyle\frac{n(n+1)}{2}$$

¿Can we elevate this closed-form solution to its second power?

Why wouldn't we be able to? We get the expression in ImCheetos reply. Is there something im missing?

 

[John3509]

It's the discrete solutions to tables 1 and 2, plus 4 other tables without using the ∑ function. Unfortunately, like the OP, it just gives me a generalized pattern for higher powered solutions. I had to cheat to get the first few patterns and then use curve fitting to get the last.

Columns y1 and y3 are the solutions to the OP's tables 1 & 2:
for table 1: $y = \frac 1 2 x^2 + \frac 1 2 x$
for table 2: $y = \frac 1 4 x^4 + \frac 1 2 x^3 + \frac 1 4 x^2$

I suppose it might have been clearer had I transposed the axes in my table.

And as far as my cheating goes, I simply plotted the numbers for each table of powers in a spreadsheet and told the spreadsheet to come up with a polynomial solution, which it is able to do for powers up to 6. But by that point a pattern had exposed itself, so getting an equation that fit for my 'y6' which required an 'x^7' just required a bit of curve fitting.

for a given $x^z$ table, $y = \frac 1 z x^{z+1} + \frac 1 2 x^z + \frac {z-1} {12} x^{z-1}$ works for $x^2$ & $x^3$.
This pattern continues for higher powers except that new factors following different patterns are added.
As to how to solve this problem without the non-methods of the OP and myself, I'll sit and wait for more clues.

Besides the equations you got for table 1 and 2, I don't get any of this.

Isn't mcastillo356s method the "proper" method to solve these types of things dealing with sequences and series, the way presented to students?

 

[mcastillo356]

Don't know if I missed the forums rules, ie, explicitly talk about solutions, especially where I haven't got anyone

 

[OmCheeto]

Don't know if I missed the forums rules, ie, explicitly talk about solutions, especially where I haven't got anyone

Well, the OP posted the solutions to the problem, but like me used a non-approved method to get them.

Besides the equations you got for table 1 and 2, I don't get any of this.

I was looking for a universal solution to this problem, where I could obtain a solution to a table which utilized any power, beyond just 1 and 3. Probably overthinking on my part. My apologies.

Isn't mcastillo356s method the "proper" method to solve these types of things dealing with sequences and series, the way presented to students?

You implied that you wanted a method that didn't involve summations. But now it's ok? In any event, it looks as though your question has been answered, so I'll unsubscribe from the thread. But thank you for posting the question, as it took me on a toboggan ride of maths with all manner of crashes and excitement.