Given that $\displaystyle \sum_{r = 1}^{n}U_r=3n^2+4n$, find $\displaystyle U_n$ and $\displaystyle \sum_{r = n+1}^{2n}U_r$
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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.
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
Very good. Now what is $\displaystyle U_{2n}-U_n~?$
9n^2+4n
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