Sigma Lh=Rh: We have Sigma notation k=3 to n; (2k-1)n....we have to prove this is equal to n^3-4n

urimagic said:
View attachment 35301
I really am trying....I just don't get this...
Click to expand...
P is incorrect. The latter one is correct. That's good progress so far.

Let's focus on the \(\displaystyle \sum_{k=3}^n 1\)

From the link, you learned that \(\displaystyle \sum_{k=1}^n c = nc\). Notice that the rule's sum starts at [imath]k=1[/imath], whereas yours start at 3 so we need to make some adjustment.

[math]\sum_{k=1}^n 1 = 1 + 1 + \sum_{k=3}^n 1\\ \sum_{k=1}^n 1 =2 + \sum_{k=3}^n 1\\ n =2 + \sum_{k=3}^n 1\\ \text{therefore } \sum_{k=3}^n 1 = n-2[/math]

Similarly, [math]\sum_{k=1}^n 2k = 2 + 4 +\sum_{k=3}^n 2k \\ 2\red{\sum_{k=1}^n k} = 6 +\sum_{k=3}^n 2k \\ 2\red{\dfrac{n(n+1)}{2}} = 6 +\sum_{k=3}^n 2k \\ \text{therefore } \sum_{k=3}^n 2k = n(n+1) -6[/math]
The sum in red is something that wasn't mentioned in the link but here's some information on it. Can you put it all together now?
 
Last edited:
BigBeachBanana said:
P is incorrect. The latter one is correct. That's good progress so far.

Let's focus on the \(\displaystyle \sum_{k=3}^n 1\)

From the link, you learned that \(\displaystyle \sum_{k=1}^n c = nc\). Notice that the rule's sum starts at [imath]k=1[/imath], whereas yours start at 3 so we need to make some adjustment.

[math]\sum_{k=1}^n 1 = 1 + 1 + \sum_{k=3}^n 1\\ \sum_{k=1}^n 1 =2 + \sum_{k=3}^n 1\\ n =2 + \sum_{k=3}^n 1\\ \text{therefore } \sum_{k=3}^n 1 = n-2[/math]

Similarly, [math]\sum_{k=1}^n 2k = 2 + 4 +\sum_{k=3}^n 2k \\ 2\red{\sum_{k=1}^n k} = 6 +\sum_{k=3}^n 2k \\ 2\red{\dfrac{n(n+1)}{2}} = 6 +\sum_{k=3}^n 2k \\ \text{therefore } \sum_{k=3}^n 2k = n(n+1) -6[/math]
The sum in red is something that wasn't mentioned in the link but here's some information on it. Can you put it all together now?
Click to expand...
Hi BigBeachBanana,

Below a question on the adjustment we had to make:IMG20230325092435.jpg
 
urimagic said:
Hi BigBeachBanana,

Below a question on the adjustment we had to make:View attachment 35331
Click to expand...
Recall
[math] \begin{aligned} \sum_{k=1}^n2k &= 2(1) + 2(2) + 2(3) + ... + 2(n)\\ &= 2 + 4 + \sum_{k=3}^n2k \end{aligned} [/math]
Similarly,
[math]\sum_{k=1}^n3k = 3(1) + 3(2) + 3(3) + ... + 3(n)[/math]
The k gets enumerated by 1 every time.

\(\displaystyle \sum_{k=1}^n 1\) does not have k inside so there's nothing to enumerate.
 
Last edited:
Then here my sum....I am beginning to understand the rules we have with regards to Sigma, just like we have Log rules, exponential rules, root rules, and so on...
However, subtracting the two expressions from one another, still does not provide the correct answer...
BigBeachBanana said:
Recall
[math] \begin{aligned} \sum_{k=1}^n2k &= 2(1) + 2(2) + 2(3) + ... + 2(n)\\ &= 2 + 4 + \sum_{k=3}^n2k \end{aligned} [/math]
Similarly,
[math]\sum_{k=1}^n3k = 3(1) + 3(2) + 3(3) + ... + 3(n)[/math]
The k gets enumerated by 1 every time.

\(\displaystyle \sum_{k=1}^n 1\) does not have k inside so there's nothing to enumerate.
Click to expand...
Understood, thank you
 

Attachments

  • IMG20230325152434[1].jpg
    IMG20230325152434[1].jpg
    3 MB · Views: 6
urimagic said:
Then here my sum....I am beginning to understand the rules we have with regards to Sigma, just like we have Log rules, exponential rules, root rules, and so on...
However, subtracting the two expressions from one another, still does not provide the correct answer...

Understood, thank you
Click to expand...
Screen Shot 2023-03-25 at 9.39.24 AM.png
You forgot the [imath]n[/imath] you factored out in the very first step.

[imath]n(n^2-4) = n^3-4n[/imath]
 
urimagic said:
AAAhhhh...Thank you so much. I have had a rough time with this one, but I have learnt a great deal...I am very grateful, thank you very much for your patience and time with me...Stay blessed.
Click to expand...
Have a look at THIS LINK!
 
You must log in or register to reply here.
Top