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Summation question

Screen Shot 2016-04-11 at 20.53.58.png

I got part a) correct, the answer being 1/(r+3) - 1/(r+4)

I was never really taught how to deal with the summation operator and would appreciate if someone could explain this to me. Thank you in advance for any help.
Reply 1
Avatar for Zacken
Zacken
22
Original post by ThomsonM98
Screen Shot 2016-04-11 at 20.53.58.png

I got part a) correct, the answer being 1/(r+3) - 1/(r+4)

I was never really taught how to deal with the summation operator and would appreciate if someone could explain this to me. Thank you in advance for any help.


What the summation operator does:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\sum_{r=1}^n f(r) = f(1) + f(2) + f(3) + \cdots + f(n)\end{equation*}



So:

Unparseable latex formula:

\displaystyle [br]\begin{align*}\sum_{r=1}^{n} \left(\frac{1}{r+3} - \frac{1}{r+4}\right) & = \frac{1}{4} - \frac{1}{5} \\ & + \frac{1}{5} - \frac{1}{6} \\ & + \frac{1}{6} - \frac{1}{7} \\ & + \quad \vdots \\ & + \frac{1}{n+2} - \frac{1}{n+3} \\ & + \frac{1}{n+3} - \frac{1}{n+4} \end{align*}



Can you see what cancels out and what you're left with?
(edited 8 years ago)
Reply 2
Avatar for Zacken
Zacken
22
Original post by ThomsonM98


I got part a) correct, the answer being 1/(r+3) - 1/(r+4)

I was never really taught how to deal with the summation operator and would appreciate if someone could explain this to me. Thank you in advance for any help.


If you're still confused, then watch this and this.
Reply 3
Avatar for B_9710
B_9710
18
Original post by ThomsonM98
Screen Shot 2016-04-11 at 20.53.58.png

I got part a) correct, the answer being 1/(r+3) - 1/(r+4)

I was never really taught how to deal with the summation operator and would appreciate if someone could explain this to me. Thank you in advance for any help.


∑r=1n(1r2+7r+12)=∑r=1n(1r+3−1r+4) \displaystyle \sum_{r=1}^{n} \left (\frac{1}{r^2+7r+12} \right )=\sum_{r=1}^{n} \left (\frac{1}{r+3}-\frac{1}{r+4} \right )

Write out the first few terms in a table or some form that makes it clear. It is a telescoping series and so you should find that a lot of the terms cancel out leaving you with a nice simple expression.
See method of differences.

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