Question regarding summation of series

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Homework Help Overview

The discussion revolves around the summation of a series defined by \(\sum^{n}_{r=1} u_r = 3n^2 + 4n\). Participants are exploring how to express \(\sum^{n-1}_{r=1} u_r\) and the implications of the given answer of \(3n^2 - 2n - 1\).

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between \(\sum^{n-1}_{r=1} u_r\) and \(\sum^{n}_{r=1} u_r\), with some attempting to derive the expression by substituting \(n\) with \(n-1\). Others question the steps taken in the derivation and seek clarification on the general term \(u_r\).

Discussion Status

There is an active exploration of different methods to arrive at the expression for \(\sum^{n-1}_{r=1} u_r\). Some participants have provided insights and alternative approaches, while others are seeking further clarification on specific steps in the reasoning process.

Contextual Notes

Participants are working within the constraints of a homework problem, which may limit the information available for deriving the general term and the summation expressions.

misogynisticfeminist
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If [tex]\sum^{n}_{r=1} u_r =3n^2 +4n[/tex], what is [tex]\sum^{n-1}_{r=1}u_r[/tex] ?

I know that [tex]\sum^{n-1}_{r=1}u_r[/tex] is equals to [tex]\sum^{n}_{r=1} u_r =3n^2 +4n - u_n[/tex] but the answer given is [tex]3n^2-2n-1[/tex]. How do i express it in that way?

thanks a lot.
 
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Use this

[tex]\sum_{r=1}^{n} r =\frac{n(n+1)}{2}[/tex]

Daniel.
 
misogynisticfeminist said:
If [tex]\sum^{n}_{r=1} u_r =3n^2 +4n[/tex], what is [tex]\sum^{n-1}_{r=1}u_r[/tex] ?

I know that [tex]\sum^{n-1}_{r=1}u_r[/tex] is equals to [tex]\sum^{n}_{r=1} u_r =3n^2 +4n - u_n[/tex] but the answer given is [tex]3n^2-2n-1[/tex]. How do i express it in that way?

thanks a lot.
Substitute n with n-1:

[tex]\sum^{n}_{r=1} u_r =3n^2 +4n[/tex]
becomes
[tex]\sum^{n-1}_{r=1} u_r =3(n-1)^2 +4(n-1)<br /> =3(n^2 -2n +1) +4(n -1)<br /> =3n^2 -2n -1[/tex]
 
Heh,again,thank god someone came with a more intuitive method...:smile:

I would have found the general term [itex]u_{r}=6r+1[/itex] ...

Daniel.
 
dextercioby said:
Use this

[tex]\sum_{r=1}^{n} r =\frac{n(n+1)}{2}[/tex]

Daniel.

hey thanks daniel. I've used that to find out that [tex]sum^{n}_{r=1} u_r[/tex] is an arithmetic progression with difference 6 and first term 7. So, i went on from there to find out the answer.

to whkoh: i don't understand your second step, [tex]\sum^{n-1}_{r=1} u_r =3(n-1)^2 +4(n-1)<br /> =3(n^2 -2n +1) +4(n -1)<br /> =3n^2 -2n -1[/tex].

would you mind explaining it to me? thanks a lot.
 
The second step is expanding out so that it becomes
[tex]3(n-1)^2+4(n-1)<br /> =3(n^2 -2n+1)+4n-4<br /> =3n^2 -6n+4n+3-4<br /> =3n^2 -2n-1[/tex]
 
I don't understand how you find the general term [tex]U_r[/tex]?
 
[tex]6 \sum_{r=1}^{n} r =6\frac{n(n+1)}{2}=3n^{2}+3n[/tex] (1)

U have [tex]3n^{2}+4n[/tex],so you need another "n"...That can be gotten noticing that

[tex]\sum_{r=1}^{n} 1 = n[/tex] (2)

Add (1) & (2) and u'll get

[tex]\sum_{r=1}^{n} \left(6r+1\right) = 3n^{2}+4n[/tex] (3)

The conclusion is simple.U found the general term,so for a summation till [itex]n-1[/itex] simply subtract the general term from the sum till [itex]n[/itex]...

Daniel.

P.S.Your method is simpler,mine is due to intuition only.
 
ahhh thanks a lot. I finally understood..
 

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