# Question regarding summation of series

1. Apr 5, 2005

### 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.

2. Apr 5, 2005

### dextercioby

Use this

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

Daniel.

3. Apr 5, 2005

### whkoh

Substitute n with n-1:

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

4. Apr 5, 2005

### dextercioby

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

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

Daniel.

5. Apr 5, 2005

### misogynisticfeminist

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, $$\sum^{n-1}_{r=1} u_r =3(n-1)^2 +4(n-1) =3(n^2 -2n +1) +4(n -1) =3n^2 -2n -1$$.

would you mind explaining it to me? thanks alot.

6. Apr 5, 2005

### whkoh

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

7. Apr 5, 2005

### whkoh

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

8. Apr 5, 2005

### 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.

9. Apr 5, 2005

### misogynisticfeminist

ahhh thanks alot. I finally understood..