1. ## summation checkin

If $\sum^{n}_{r=1}U_r=3n^2+2n$, find u .

My working :

From the result , i can see that it uses this formula

$\sum^{n}_{r=1}\frac{n(n+1)}{2}$ , so r exists.

$\frac{1}{2}xr^2+\frac{1}{2}xr-n=3n^2+2n$

By comparing , i got x=6 , and thus the answer is 6r-1 .

I just want to know if there is a simpler approach to questions like this .
Thanks .

If $\sum^{n}_{r=1}U_r\:=\:3n^2+2n$, find $U.$

Like you, I saw that the function is linear: . $U_r \:=\:ar + b$

That is, we have: . $\sum^n_{r=1}\,(ar + b) \;=\;3n^2 + 2n$

Then: . $a\sum^n_{r=1}r + b\sum^n_{r=1}1 \;=\;3n^2+2n$

Knowing some summation formulas, we have: . $a\cdot\frac{n(n+1)}{2} + b\cdot n \:=\:3n^2 + 2n$

. . which simplifies to: . $\frac{a}{2}n^2 + \left(\frac{a}{2}+b\right)n \;=\;3n^2+2$

Equating coefficients, we have: . $\begin{Bmatrix}\dfrac{a}{2} &=&3 \\ \\[-4mm] \dfrac{a}{2} + b &=& 2 \end{Bmatrix}$

. . Solve the system and get: . $a = 6,\;b = -1$

Therefore: . $U_r \:=\:6r-1$

3. THanks Soroban , your working is so much neater .

4. Simpler :

$\sum^{n}_{r=1}U_r=3n^2+2n$

$U_n = \sum^{n}_{r=1}U_r - \sum^{n-1}_{r=1}U_r$

$U_n = 3n^2+2n-3(n-1)^2-2(n-1) = 6n-1$