a simple one but I cant remember how to do this!!!
Find sum of the series...
$\displaystyle
\sum\limits_{r = 5}^{r = 90} {(3r - 2)}
$
$\displaystyle \sum_{r = 5}^{90} {(3r - 2)} $
$\displaystyle = \sum_{r = 5}^{90}3r + \sum_{r = 5}^{90}(-2)$
$\displaystyle = 3 \sum_{r = 5}^{90}r - 2 \sum_{r = 5}^{90}1$
$\displaystyle = 3 \left ( \sum_{r = 1}^{90}r - \sum_{r = 1}^4r \right ) - 2 \left ( \sum_{r = 1}^{90}1 - \sum_{r = 1}^41 \right )$
Can you take it from here?
-Dan
Hello,
Let $\displaystyle a_r=3r-2$
$\displaystyle a_{r+1}=3(r+1)-2=(3r-2)+3=a_r+{\color{red}3}$
Therefore :
$\displaystyle \sum a_r$ is an arithmetic series of progression 3.
Hmmm, can you continue ?
--------------------
Here is another method :
$\displaystyle \sum_{r=5}^{90} (3r-2)=3 \sum_{r=5}^{90} r-\sum_{r=5}^{90} 2$
You should know the general formula for $\displaystyle \sum r$.
And for $\displaystyle \sum 2$, represent yourself adding 2, adding again, and again, and again..
(let's see first if you can do the first one ^^)
Edit : oh, well...topsquark spoiled it
Hello, dankelly07!
Another approach . . .
Find sum of the series: .$\displaystyle \sum\limits_{r = 5}^{90} (3r - 2)$
We have: .$\displaystyle S \:=\:13 + 16 + 19 + 22 + \hdots + 268$
This is an arithmetic series.
. . It has: .first term $\displaystyle a = 13$, common difference $\displaystyle d = 3$, and $\displaystyle n = 86$ terms.
Its sum is: .$\displaystyle S \;=\;\frac{n}{2}\bigg[2a + (n-1)d\bigg] \;=\;\frac{86}{2}\bigg[2(13) + 85(3)\bigg] \;=\;12,083$