Finding the general term from summation

**Punch** · May 21st 2011, 06:13 AM

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

**Plato** · May 21st 2011, 06:39 AM

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.

**Punch** · May 21st 2011, 07:49 AM

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

**Plato** · May 21st 2011, 07:53 AM

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

**Duke** · May 21st 2011, 08:08 AM

9n^2+4n

Thread: Finding the general term from summation