Finding the general term from summation

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May 21st 2011, 06:13 AM
Punch

Finding the general term from summation

Given that $\sum_{r = 1}^{n}U_r=3n^2+4n$ , find $U_n$ and $\sum_{r = n+1}^{2n}U_r$
May 21st 2011, 06:39 AM
Plato

Quote:

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.
May 21st 2011, 07:49 AM
Punch

Quote:

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 :D
May 21st 2011, 07:53 AM
Plato

Very good.
Now what is $U_{2n}-U_n~?$
May 21st 2011, 08:08 AM
Duke

9n^2+4n

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