# Thread: Finding the general term from summation

1. ## Finding the general term from summation

Given that $\sum_{r = 1}^{n}U_r=3n^2+4n$, find $U_n$ and $\sum_{r = n+1}^{2n}U_r$

2. Originally Posted by Punch
Given that $\sum_{r = 1}^{n}U_r=3n^2+4n$, find $U_n$ and $\sum_{r = n+1}^{2n}U_r$
$U_n=\left(\sum_{r = 1}^{n}U_r\right)-\left(\sum_{r = 1}^{n-1}U_r\right)$.

Now you show us some effort.

3. Originally Posted by Plato
$U_n=\left(\sum_{r = 1}^{n}U_r\right)-\left(\sum_{r = 1}^{n-1}U_r\right)$.

Now you show us some effort.
$U_n=\left(\sum_{r = 1}^{n}U_r\right)-\left(\sum_{r = 1}^{n-1}U_r\right)=(3n^2+4n)-(3(n-1)^2+4(n-1))$
$=3n^2+4n-(3n^2-6n+3+4n-4)$
$=6n+1$

thanks

4. Very good.
Now what is $U_{2n}-U_n~?$

5. 9n^2+4n