Lecture 2

Faculty of Science; Department of Mathematics
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Readings For This Lecture

Keshet Course Notes:

Chapter 1, pages 1 to 13 (up to subsection 1.6)

Summary

When one wants to sum up a large set of numbers, it helps to have some sort of notation that cleans up something like $1+2+3+4+5+6+\cdots +200$ .

This notation would be what is called sigma, or summation notation. For the sum stated above, the notation would be $\sum _{k=1}^{200}k$

The number below the sigma (k=1) is called the "index of summation" and it represents what term the sum is starting from, in this case, it is the first term.

The superscript, or whatever is above the sigma shows at what point the summation stops.

Suppose that we want to sum up five 1s, or $1+1+1+1+1$ . To put this into sigma notation we would write $\sum _{k=1}^{5}1$ . This is because we want the sum of term 1 through 5 of a formula, "1". This sum, of course, would just lead to the number 5.

The general formula for this would be $\sum _{k=1}^{N}1=N$

This fact can also lead to the very useful fact that, if what you are summing up has a common factor in it, you can bring that factor out of the notation and multiply it to the sum after it is found. e.g. $\sum _{k=1}^{20}3=3\sum _{k=1}^{20}1=3\left(20\right)=60$

Lets now take a look at how combine similar sums. $\sum _{k=1}^{7}3^{k}-\sum _{k=3}^{7}3^{k}=\left(3+3^{2}+\cdots +3^{7}\right)-\left(3^{3}+3^{4}+\cdots +3^{7}\right)=3+3^{2}$ which is really just $\sum _{k=1}^{2}3^{k}$ .

Another way of simplifying sums is to expand. When you have a sum of two terms being added or subtracted (e.g. $\sum _{k=0}^{8}\left(1-2^{k}\right)$ ) you can split each part into it's own sum.

In this case, that would mean $\sum _{k=0}^{8}1-\sum _{k=0}^{8}2^{k}$ .

There are a few formulae for actually determining sums from the summation notation.

The first one we will look at is Gauss' formula which will aid us in determining the sum of $\sum _{k=1}^{N}k$ which can also be written as $1+2+\cdots +\left(N-1\right)+N$

If we add this series to itself flipped around, we would get

$S=1+2+\cdots +\left(N-1\right)+N$

+

$S=N+\left(N-1\right)+\cdots +2+1$

$2S=\left(N+1\right)+\left(N+1\right)+\cdots +\left(N+1\right)+\left(N+1\right)$

Which means there are N terms of $\left(N+1\right)$ so $2S=N\left(N+1\right)$ which means $S={\frac {N\left(N+1\right)}{2}}$

This conclusion that was just reached is Gauss' formula: $\sum _{k=1}^{N}k={\frac {N\left(N+1\right)}{2}}$

There are also formulae for adding up N terms of $k^{2}$ and $k^{3}$

These are:

$\sum _{k=1}^{N}k^{2}={\frac {N\left(N+1\right)\left(2N+1\right)}{6}}$

and

$\sum _{k=1}^{N}k^{3}=\left({\frac {N\left(N+1\right)}{2}}\right)^{2}$

Exercises

Reference the appropriate formula from the course formula sheet when required.

1. Write $1+2+4+6+8+10+\cdots +196$ in summation notation.

Solution
$1+\left(\sum _{k=1}^{98}2k\right)$

2. Write $1+{\tfrac {1}{2}}+{\tfrac {1}{9}}+{\tfrac {1}{16}}+{\tfrac {1}{25}}+\cdots$ in summation notation.

Solution
${\frac {1}{4}}+\left(\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\right)$

3. Simplify $\sum _{k=5}^{\infty }\left({\frac {1}{2}}\right)^{k}+\sum _{k=2}^{4}\left({\frac {1}{2}}\right)^{k}$ .

Solution
$\sum _{k=2}^{\infty }\left({\frac {1}{2}}\right)^{k}$

4. Simplify $2\sum _{y=0}^{100}y+\sum _{y=1}^{100}y^{2}+\sum _{y=0}^{100}1$ .

Solution
$1+\left(\sum _{y=1}^{100}\left(y^{2}+2y+1\right)\right)$

5. Show that the sequences $\sum _{m=0}^{10}(m+1)^{2}$ and $\sum _{n=1}^{11}n^{2}$ are equivalent.

Solution
This uses the facts that $\sum _{k=1}^{N}k={\frac {N\left(N+1\right)}{2}}$ and $\sum _{k=1}^{N}k^{2}={\frac {N\left(N+1\right)\left(2N+1\right)}{6}}$ . First, we can expand $\sum _{m=0}^{10}(m+1)^{2}$ to be $\sum _{m=0}^{10}m^{2}+2m+1$ and split that into $\sum _{m=0}^{10}m^{2}+2\sum _{m=0}^{10}m+\sum _{m=0}^{10}1$ which, when considering term 0 in the first two sums is equivalent to $\sum _{m=1}^{10}m^{2}+2\sum _{m=1}^{10}m+\sum _{m=0}^{10}1$ . We then put the corresponding summations into the proper formulae to get ${\frac {10\left(11\right)\left(21\right)}{6}}+{\frac {2\left(10\right)\left(11\right)}{2}}+11=506$ The second part, $\sum _{n=1}^{11}n^{2}$ can simply be computed with the second formula to be ${\frac {11\left(12\right)\left(23\right)}{6}}=506$ We can see that both summation notations added up to the same value of 506

6. Compute the sum $\sum _{n=10}^{60}n$ .

Solution
Using the fact that $\sum _{k=1}^{N}k={\frac {N(N+1)}{2}}$ $\sum _{k=1}^{60}k-\sum _{k=1}^{9}k={\frac {60(61)}{2}}-{\frac {9(10)}{2}}=1785$

7. Compute the sum $\sum _{k=2}^{50}(k^{2}-2k+1)$ .

Solution
This uses the facts that $\sum _{k=1}^{N}k={\frac {N\left(N+1\right)}{2}}$ and $\sum _{k=1}^{N}k^{2}={\frac {N\left(N+1\right)\left(2N+1\right)}{6}}$ . We can first expand the sum to include the first term and subtract it in a further step, $\left(\sum _{k=1}^{50}(k^{2}-2k+1)\right)-\left(\sum _{k=1}^{1}(k^{2}-2k+1)\right)$ . This then can be expanded and simplified to $\left(\sum _{k=1}^{50}k^{2}-2\sum _{k=1}^{50}k+\sum _{k=1}^{50}1\right)-\left(1^{2}-2\left(1\right)+1\right)$ This, when put into the proper formulae comes out to $\left({\frac {50\left(51\right)\left(101\right)}{6}}-\left({\frac {2\left(50\right)\left(51\right)}{2}}\right)+50\right)-\left(1-2+1\right)=40425-0=40425$

8. Find the sum of the first 50 odd integers, $1+3+5+\cdots$ .

Solution
Using the fact that $\sum _{k=1}^{N}k={\frac {N(N+1)}{2}}$ $\sum _{k=0}^{49}1+2k=\sum _{k=0}^{49}1+2\sum _{k=0}^{49}k=50+{\frac {2\left(49\right)\left(50\right)}{2}}=2500$

9. Compute the following sum: $\sum _{i=1}^{12}i(1-i)+2^{i}$ .

Solution
This uses $\sum _{k=1}^{N}k={\frac {N\left(N+1\right)}{2}}$ , $\sum _{k=1}^{N}k^{2}={\frac {N\left(N+1\right)\left(2N+1\right)}{6}}$ and $\sum _{k=0}^{N}r^{k}={\frac {1-r^{N+1}}{1-r}}$ . The sum can first be expanded to $\sum _{i=1}^{12}i-1^{2}+2^{i}.$ This then can be broken up into $\sum _{i=1}^{12}i-\sum _{i=1}^{12}i^{2}+\sum _{i=1}^{12}2^{i}$ Putting the proper values into the formulae you get ${\frac {12\left(13\right)}{2}}-{\frac {12\left(13\right)\left(25\right)}{6}}+{\frac {1-2^{13}}{-1}}-1$ The -1 was at the end because the formula that was used uses a sum that goes from term 0 to N. In our case we didn't have a term 0 so we had to subtract what that term would have added on, which is $2^{0}=1$ . Therefore, the sum comes out to a value of 7618.

10. A triangular pyramid of spheres has one sphere on the top layer, 2+1 = 3 spheres on the second layer, 3+2+1=6 spheres on the third layer, and so on. Write down a formula for the number of spheres needed for a pyramid that is $N$ layers deep. How many spheres are needed for a pyramid that is 100 layers deep?

Solution
This uses $\sum _{k=1}^{N}k={\frac {N\left(N+1\right)}{2}}$ and $\sum _{k=1}^{N}k^{2}={\frac {N\left(N+1\right)\left(2N+1\right)}{6}}$ Formula: $\sum _{k=1}^{N}{\frac {k\left(k+1\right)}{2}}={\frac {1}{2}}\sum _{k=1}^{N}k^{2}+k={\frac {1}{2}}\left(\sum _{k=1}^{N}k^{2}+\sum _{k=1}^{N}k\right)$ For a pyramid 100 layers deep, we plug the proper values into the formulae and get: ${\frac {1}{2}}\left(\sum _{k=1}^{100}k^{2}+\sum _{k=1}^{100}k\right)={\frac {1}{2}}\left({\frac {100\left(101\right)\left(201\right)}{6}}+{\frac {100\left(101\right)}{2}}\right)=171700$ spheres are needed.