Question regarding summation of series

[misogynisticfeminist]

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

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

thanks alot.

 

[dextercioby]

Use this

\sum_{r=1}^{n} r =\frac{n(n+1)}{2}

Daniel.

 

[whkoh]

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

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

thanks alot.

Substitute n with n-1:

\sum^{n}_{r=1} u_r =3n^2 +4n
becomes
\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

 

[dextercioby]

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

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

Daniel.

 

[misogynisticfeminist]

Use this

\sum_{r=1}^{n} r =\frac{n(n+1)}{2}

Daniel.

hey thanks daniel. I've used that to find out that sum^{n}_{r=1} u_r 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, \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.

would you mind explaining it to me? thanks alot.

 

[whkoh]

The second step is expanding out so that it becomes
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<br />

 

[whkoh]

I don't understand how you find the general term U_r?

 

[dextercioby]

6 \sum_{r=1}^{n} r =6\frac{n(n+1)}{2}=3n^{2}+3n (1)

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

\sum_{r=1}^{n} 1 = n (2)

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

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

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

Daniel.

P.S.Your method is simpler,mine is due to intuition only.

 

[misogynisticfeminist]

ahhh thanks alot. I finally understood..