Question regarding summation of series

[misogynisticfeminist]

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 alot.

 

[dextercioby]

Use this

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

Daniel.

 

[whkoh]

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 alot.

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)
=3(n^2 -2n +1) +4(n -1)
=3n^2 -2n -1[/tex]

 

[dextercioby]

Heh,again,thank god someone came with a more intuitive method...

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

Daniel.

 

[misogynisticfeminist]

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)
=3(n^2 -2n +1) +4(n -1)
=3n^2 -2n -1[/tex].

would you mind explaining it to me? thanks alot.

 

[whkoh]

The second step is expanding out so that it becomes
[tex]3(n-1)^2+4(n-1)
=3(n^2 -2n+1)+4n-4
=3n^2 -6n+4n+3-4
=3n^2 -2n-1
[/tex]

 

[whkoh]

I don't understand how you find the general term [tex]U_r[/tex]?

 

[dextercioby]

[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.

 

[misogynisticfeminist]

ahhh thanks alot. I finally understood..