# Finding the general term from summation

• May 21st 2011, 06:13 AM
Punch
Finding the general term from summation
Given that $\displaystyle \sum_{r = 1}^{n}U_r=3n^2+4n$, find $\displaystyle U_n$ and $\displaystyle \sum_{r = n+1}^{2n}U_r$
• May 21st 2011, 06:39 AM
Plato
Quote:

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

$\displaystyle U_n=\left(\sum_{r = 1}^{n}U_r\right)-\left(\sum_{r = 1}^{n-1}U_r\right)$.

Now you show us some effort.
• May 21st 2011, 07:49 AM
Punch
Quote:

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

Now you show us some effort.

$\displaystyle U_n=\left(\sum_{r = 1}^{n}U_r\right)-\left(\sum_{r = 1}^{n-1}U_r\right)=(3n^2+4n)-(3(n-1)^2+4(n-1))$
$\displaystyle =3n^2+4n-(3n^2-6n+3+4n-4)$
$\displaystyle =6n+1$

thanks :D
• May 21st 2011, 07:53 AM
Plato
Very good.
Now what is $\displaystyle U_{2n}-U_n~?$
• May 21st 2011, 08:08 AM
Duke
9n^2+4n