Question

How to show that the sum from k=1 to n of k^3 is equivalent to ((n(n+1))/ 2)^2 ? I'm looking for clues, not your full answer in complete, thank you!

Answer 1

Have you learned induction yet?

Answer 2

Yes, I'm somewhat familiar.

Answer 3

Well, that's all you have to use here.

Answer 4

I need to show that it is true using the telescoping method for this assignment.

Answer 5

Your formula is wrong. Double check..

Answer 6

Ok,hmmm

Answer 7

Hint :
\[\sum\limits_{k=1}^n [k^3 - (k-1)^3] = n^3\]

Answer 8

The formula is alright. \(\left(\frac{n(n+1)}{2}\right)^2 = \frac{n^2(n+1)^2}{4}\)

Did you succeed with the telescoping?

Answer 9

Actually, in \(k^3-(k-1)^3\) , there are no terms \(k^3\). While there is if you consider \(k^4-(k-1)^4\).
So start with the sum ganeshie8 proposed, with power \(4\), and the RHS is \(n^4\).

Answer 10

\( \sum_{k=1}^n \bigl(k^4 - (k^4 - 4k^3 + 6k^2 - 4k + 1)\bigr) =n^4\).

Answer 11

There's something about the telescoping that maybe I'm not getting yet.

For k^3 to be a telescoping series form \[\sum_{i=1}^{n}(A _{i}-A _{i+1})\] One may deduce that that it's terms will be (1^3 -2^3 ) + (2^3 - 3^3)... + ((n-1)^3-n^3)
Otherwise expressed...not as what I must prove..

Answer 12

How can induction help if I need prove it by the telescoping method?

Answer 13

The telescoping sum is \(\sum_{i=1}^n A_{i}-A_{i-1} = A_n\) (with \(A_0=0\)).
So, whatever \(A_k\), the telescoping is always equal to \(A_n\). Now the reason to choose \(A_i=i^4\) is because \(A_i - A_{i-1} = c i^3 + c' i^2 + c'' i + c'''\). You must know the sum of the \(i^2\) and the sum of the \(i\). The sum of the \(i^3\) is the unknown in this equation\[\sum_i ci^3 + \sum_ic'i^2 +\sum_i c''i + \sum_i c''' = n^4\]

You will not need induction here.

Answer 14

So that
\[ c\sum_{i=1}^n i^3 = n^4 - \biggl( c'\sum_ii^2 +c''\sum_i i + c'''\sum_i 1 \biggr) \]

Answer 15

Telescoping doesn't require induction, can be shorter, but you need to come up with a trick.

To summarize, to compute \(\sum_{i=1}^n t_i\), you have to come up with another sum: \(\sum_{i=1}^n u_i\), such that \(u_i-u_{i-1}\) contains \(t_i\).
You don't just write \(\sum_{i=1}^n (t_i-t_{i-1})\).

Answer 16

@daniel.ohearn1 you have to determine the constants \(c,c',c'',c'''\) (by Newton) and replace \(\sum_{i=1}^n i\) by \(\frac{n(n+1)}{2}\), for example.
Can you go through it?

Answer 17

Oh that's by Newton? I think I should wait until I can have more light tomorrow. I would get 1-n^3 writing out the terms in the sum of k^3 from k=1 to n. But to show 1-n^3 = what I'm trying show, well it's gonzo, I just don't know yet..

Answer 18

By "Newton", I mean: \(k^4-(k-1)^4 = k^4 - (k^4 -4k^3 + 6k^2 - 4k + 1)\)
And you don't want to find a telescoping sum equal to "what im trying to show", but you want to see "what im trying to show" as part of the telescoping sum:
\[\sum_{k=1}^n k^4 - (k-1)^4 = \sum_{k=1}^n (\text{\(4k^3\) and other simple things}) = \text{knownexpression}.\]

Answer 19

I feel like adding an extra bit of notation is good for telescoping series,

\[\Delta f(k) = f(k)-f(k-1)\]
Then this means you have exactly the finite difference version of the 'fundamental theorem of calculus' which is called the hokey name of 'telescoping series.

\[\sum_{k=1}^n \Delta f(k) = f(n)-f(0)\]

Now so what if this doesn't help you right? So let's use it.

\[\sum_{k=1}^nk^3 = \left(\frac{n(n+1)}{ 2}\right)^2 \]
Accordingly, we must then have:

\[\Delta f(k) = k^3\] and \[f(n)-f(0) = \left(\frac{n(n+1)}{ 2}\right)^2 \]
Since f(0) is a constant we can immediately test to see if these are the same function,

\[\Delta [f(n)-f(0)] = \Delta f(n) = \left(\frac{n(n+1)}{ 2}\right)^2 -\left(\frac{(n-1)n}{ 2}\right)^2 \]
Upon simplification of the RHS you should get:
\[\Delta f(n) =n^3\]

So I'll do that:
\[ \left(\frac{n(n+1)}{ 2}\right)^2 -\left(\frac{(n-1)n}{ 2}\right)^2 = \frac{n^2}{4}[(n+1)^2-(n-1)^2]\]\[ \frac{n^2}{4}[(n+1)^2-(n-1)^2] = \frac{n^2}{4} [n^2+2n+1-n^2+2n-1]=\frac{n^2}{4} 4n=n^3\]

Which confirms it. This method is pretty mindless too for problems like these once you know how it works... If you don't for some reason like this cause it's not proper for your class or something, then maybe you can find a way to turn it into something uglier... :P

Answer 20

This is what I have, does this look complete?

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

Then by the telescoping method

\[\sum_{k=1}^{n}\Delta f(k) = \Delta \left( \left( \frac{ n(n+1) }{ 2 } \right)^2 - \left( \frac{n( n-1) }{ 2 } \right)^2 \right)\]

Answer 21

\[\rightarrow \Delta \left( \frac{ n^4+2n^3+n^2-(n^2-n)^2 }{ 4 } \right)\]

Answer 22

\[\rightarrow \Delta \left( \frac{ n^4+2n^3+n^2-n^4+2n^3-n^2 }{ 4} \right) = \Delta n^3 \]

Answer 23

And the change of n^3 can also be written
\[\sum_{k=1}^{n}\Delta f(k) \]

Which we know to be the original sum:

\[\sum_{k=1}^{n}k^3\]

Answer 24

Also

\[\Delta n^3 = \int\limits_{1}^{n}x^3dx = \sum_{x=1}^{n}x^3\Delta x = \sum_{k=1}^{n}k^3\]

Where \[n \epsilon \mathbb{R}, n \ge 1 \]

Answer 25

And delta k is implied in the definition of the Reimann Sum, right?

Answer 26

@daniel.ohearn1 I don't understand what you're asking.
What @Kainui said is that if you want to \(\verb|check|\) if \(f(n)\) is the formula for \(\sum_{k=1}^n k^3\), you can just take the difference of \(f(n)\) and \(f(n-1)\):
\[ f(n) - f(n-1) \stackrel{\text{should be}}{=} \sum_{k=1}^n k^3 - \sum_{k=1}^{n-1} k^3 = n^3 \]
He also showed you all the steps.

Answer 27

+ check \(f(1)=1\).

Answer 28

Still good to support my proof, thanks!

Answer 29

It really does. If \(f(n)-f(n-1) = n^3\), then what is \(f(n)\)?
Here: \(f(n) = (f(n)-f(n-1)) + (f(n-1)-f(n-2)) + \dotsb + (f(2)-f(1)) + (f(1) - 0)\). (which is the telescoping sum).

And \( (f(n)-f(n-1)) + (f(n-1)-f(n-2)) + \dotsb + (f(2)-f(1)) + (f(1) - 0)\) is actually the same (term by term) as
\( n^3 + (n-1)^3 + \dotsb + 1^3\).

So it's proven.