Write the indicated sum in sigma notation:
1-(1/2)+(1/3)-(1/4)+...-(1/100)
Another one using Special Sum Formulas
nEi=1 (2i-3)^2
I wanted to break it up but the square is throwing me off
btw the n is on top of the E and i=2 is below it.
I think I understand the basic concept of it, however I don't seem to have a full grasp of it once it gets a little complex. I know what each variable does... however something like what you did for my problem seems confusing. I don't see how your solution solves my problemOriginally Posted by ThePerfectHacker
You start from one and you have,Originally Posted by pakman
$\displaystyle (-1)^{1-1}\frac{1}{1}=(-1)^0\frac{1}{1}=1/1$---> first term good
$\displaystyle (-1)^{2-1}\frac{1}{2}=(-1)^1\frac{1}{2}=-1/2$---> second term good
$\displaystyle (-1)^{3-1}\frac{1}{3}=(-1)^2\frac{1}{3}=1/3$---> thrid term good.
And so on
Hello, pakman!
Another one using Special Sum Formulas:
. . $\displaystyle \sum^n_{i=1} (2i-3)^2$
I'm only guessing, but are these the Special Sum Formulas?
. . $\displaystyle \sum^n_{i=1}1\;=\;n \qquad \sum^n_{i=1}i\;=\;\frac{n(n+1)}{2} \qquad \sum^n_{i=1}i^2\;=\;\frac{i(i+1)(2i+1)}{6}$
We have: .$\displaystyle \sum^n_{i=1}(2i-3)^2\;=\;\sum^n_{i=1}\left(4i^2 - 12i + 9\right) $
. . Then: . . . . $\displaystyle 4\sum^n_{i=1}i^2 \quad\;\;- \quad\;\;12\sum^n_{i=1}i \quad+ \quad9\sum^n_{i=1}1 $
. . . . . . . . . . . . $\displaystyle \downarrow$ . . . . . . . . . . $\displaystyle \downarrow$ . . . . . . . $\displaystyle \downarrow$
. . $\displaystyle = \;4\overbrace{\left[\frac{n(n+1)(2n+1)}{6}\right]} - 12\overbrace{\left[\frac{n(n+1)}{2}\right]} \;+ \;9\overbrace{[n]} $
which simplifies to: .$\displaystyle \frac{n}{3}\left(4n^2 - 12n + 11)$