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Assuming we don't have a calculator that can do summation notation. My class is not up to summation yet, but I'm asking a question involving this concept because I'm not all that experienced using it. My calculator can do summation but that's because I programmed it to. But when we actually catch up to this concept, are we to calculate it by hand?

Take, for example:

$\displaystyle \sum_{i = 0}^3 2i + 1$

Do you think they will tell us for the $4$ values of $i$ (from $0$ to $3$), do the following operation, or is there some other faster way of doing it?

Sorry if this is a silly question, this has just been on my mind...

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With a small sum like that you can even do it in your head. Depending on just what’s in your course, you may learn some formulas for evaluating certain kinds of sums quickly, including this one, but in general yes, you’ll have to evaluate them by hand. –  Brian M. Scott Apr 6 '12 at 0:27
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Did you mean $\sum_{i=0}^3 (2i+1)$ or $(\sum_{i=0}^3 2i)+1$? –  user26770 Apr 6 '12 at 1:02
    
The first one. :) –  David Apr 6 '12 at 11:52

1 Answer 1

up vote 9 down vote accepted

Ask your teacher. Mileage may vary. Just make sure you know you have the right answer, whichever method you use. In this case, there is actually a very nice formula: $$\sum_{k=0}^{n-1}{2k+1}=n^2$$ and in your case, $n=4$. You can also think of this as $4=3-0+1$ (the number of terms being added) times the average value of each term, which is $2x+1$ for $x=\frac{0+3}{2}$. And the formula works because $(n+1)^2=n^2+(2n+1)$. I have formulated the summation going from $0$ to $n-1$ because this has $n$ terms. Perhaps you would be more comfortable with the $(n+1)$-term version: $$\sum_{k=0}^{n}{2k+1}=(n+1)^2$$ Or perhaps you could generalize this to $$\sum_{k=a}^{b}{2k+1}=(b+1)^2-a^2=(b+a+1)(b-a+1)$$ for $a\le b$. It takes a bit of getting used to, and even then a bit of care working with summations, but it is time very well spent, in my opinion. In fact, there are some really interesting generalizations: $$\sum_{i=1}^{n}\,i=\frac{n(n+1)}{2}$$ $$\sum_{i=1}^{n}\,\frac{i(i+1)}{2}=\frac{n(n+1)(n+2)}{1\cdot2\cdot3}$$ and some other ones I'm sure you'll love: $$\sum_{i=0}^{n-1}\,a^i=\frac{a^n-1}{a-1}\qquad\text{for }a\ne1$$ $$\sum_{n=1}^\infty\,\frac1{n^2}=\frac{\pi^2}{6}$$ More in the "Capital-sigma notation" section of the Wikipedia article with the same name. But some are not as easy as one might hope:

  1. $$\sum_{i=0}^n\,i^k=$$

  2. general formula for arbitrary $k$: (hard)

  3. $$\sum_{n=1}^\infty\,\frac1{n^s}=$$ (a very important but mysterious function)

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