\[ \sum _{n=-1}^{4} \sqrt {n+1} =\sqrt {0}+\sqrt {\answer [given]{1}}+\sqrt {\answer [given]{2}}+\sqrt {\answer [given]{3}}+\sqrt {\answer [given]{4}}+\sqrt {\answer [given]{5}}\]
We introduce sigma notation.
1 The notation itself
Sigma notation is a way of writing a sum of many terms, in a concise form. A sum in sigma notation looks something like this:
\[ \sum _{k=1}^{5} 3k \]
The \(\Sigma \) (sigma) indicates that a sum is being taken. The variable \(k\) is called the index of the sum. The numbers at the top and bottom of the \(\Sigma \) are called the upper and lower limits of the summation. In this case, the upper limit is \(5\), and the lower limit is \(1\). The notation means that we will take every integer value of \(k\) between \(1\) and \(5\) (so \(1\), \(2\), \(3\), \(4\), and \(5\)) and plug them each into the summand formula (here that formula is \(3k\)). Then those are all added together.
\[ \sum _{k=1}^{5} 3k= 3\cdot 1 + 3\cdot 2 + 3\cdot 3 + 3\cdot 4 + 3\cdot 5 = 45 \]
Write out what is meant by the following:
\[ \sum _{k=0}^{3} \frac {1}{k+1} \]
Here, the index \(k\) takes the
values \(0\), \(1\), \(2\), and \(3\). We’ll plug those each into \(\frac {1}{k+1}\) and add them together.
\[ \sum _{k=0}^3 \frac {1}{k+1} = \frac {1}{\answer [given]{0}+1} + \frac {1}{\answer [given]{1}+1} + \frac {1}{\answer [given]{2}+1} + \frac {1}{\answer [given]{3}+1} \]
Write out what is meant by the following:
\[ \sum _{i=1}^{8} \left (-1\right )^i \]
The index variable here is
written as \(i\) instead of \(k\). That’s ok. The most common variables to use for
indexes include \(i\), \(j\), \(k\), \(m\), and \(n\). \begin{align*} \sum _{i=1}^8 (-1)^i = (-1)^{\answer [given]{1}} &+ (-1)^{\answer [given]{2}} + (-1)^{\answer [given]{3}} + (-1)^{\answer [given]{4}} \\ &+ (-1)^{\answer [given]{5} }+ (-1)^{\answer [given]{6}} + (-1)^{\answer [given]{7}} + (-1)^{\answer [given]{8}} \end{align*}
This is equal to: \begin{align*} &= -1 + 1 + -1 + 1 + -1 + 1 + -1 + 1\\ &= \answer [given]{0} \end{align*}
Try one on your own.
Write out what is meant by the following (no
need to simplify):
Let’s try going the other way around.
Write the following sum in sigma
notation.
\[ 2 + 4 + 6 + 8 + \ldots + 22 + 24 \]
Notice that we can factor a \(2\) out of each term to rewrite this
sum as
\[ \answer [given]{2}\cdot 1 + \answer [given]{2} \cdot 2 + \answer [given]{2}\cdot 3 + \answer [given]{2} \cdot 4 + \ldots + \answer [given]{2}\cdot 11 + \answer [given]{2}\cdot 12 \]
That means that we are adding together \(\answer [given]{2}\) times every number
between \(1\) and \(12\). If we use \(k\) as our index, the sigma notation could be \[ \sum _{k=1}^{12} \answer [given]{2k} \]
There is no need to use \(k\) as our index variable. We could have just as
easily used \(m\) or \(j\) instead. \[ \sum _{k=1}^{12} 2k = \sum _{m=1}^{12} 2m = \sum _{j=1}^{12} 2j\]
Notice, that these are NOT the same as \[ \sum _{k=1}^{12} 2m \]
Write the following sum in sigma notation.
\[ 1 - \frac {1}{2} + \frac {1}{4} - \frac {1}{8} + \ldots - \frac {1}{64} + \frac {1}{128} \]
This one is a little more
complicated. We’ll worry about the signs later, first we’ll deal with the
numbers themselves. Do you notice a pattern in the terms? Sure, we
get from one term to the next term by dividing by 2. That is:
\begin{align*} 1 &= \frac {1}{2^0} = \left (\frac {1}{2}\right )^0\\ \frac {1}{2} &= \frac {1}{2^1} = \left (\frac {1}{2}\right )^{1}\\ \frac {1}{4} &= \frac {1}{2^2} = \left (\frac {1}{2}\right )^{2}\\ &\vdots \\ \frac {1}{128} &= \frac {1}{2^7} = \left (\frac {1}{2}\right )^{7} \end{align*}
If we call our index variable \(k\), then \(k\) should go from \(0\) to \(\answer [given]{7}\), and the numbers themselves are just \(\left (\frac {1}{2}\right )^{\answer [given]{k}}\). Now we need to deal with the signs. We say above that \((-1)^k\) will alternate between \(+1\) and \(-1\). That means, if we multiply the terms we just found by \((-1)^k\), they will alternate between \(+\) and \(-\). We are starting with \(k=0\), so \((-1)^0 = +1\) will give us the alternation starting at the sign we want.
\[ \sum _{k=0}^{7} (-1)^k \left (\frac {1}{2}\right )^{k} = \sum _{k=0}^7 \left ( \answer [given]{\frac {-1}{2}} \right )^k \]
Try one on your own.
Write the following sum in sigma notation.
\[ \frac {1}{\sqrt {2}} - \frac {2}{\sqrt {3}} + \frac {3}{\sqrt {4}} - \frac {4}{\sqrt {5}}+ \ldots + \frac {51}{\sqrt {52}} - \frac {52}{\sqrt {53}}\]
We
write: \[ \sum _{k=2}^{53} \left ( \answer [given]{(-1)^{k}\frac {k-1}{\sqrt {k}} } \right ) \]
2 Calculating with sigma notation
We want to use sigma notation to simplify our calculations. To do that, we will need to know some basic sums. First, let’s talk about the sum of a constant. (Notice here, that our upper limit of summation is \(n\). \(n\) is not the index variable, here, but the highest value that the index variable will take.)
\[ \sum _{k=1}^n C = \underbrace {C+C+ \dots +C}_{\text {n terms}} \]
This is a sum of \(n\) terms, each of them having a value \(C\). That is, we are adding \(n\) copies of \(C\). This sum is just \(nC\). The other basic sums that we need are much more complicated to derive. Rather than explaining where they come from, we’ll just give you a list of the final formulas, that you can use.
Now that we have this list, let’s use them to compute.
Because sigma notation is just a new way of writing addition, the usual properties of addition still apply, but a couple of the important ones look a little different.
- Commutativity and Associativity:
-
\[ \sum _{k=1}^{n} (a_k + b_k) = \sum _{k=1}^n a_k + \sum _{k=1}^n b_k \]
- Distribution:
-
\[ \sum _{k=1}^{n} c \cdot a_k = c \sum _{k=1}^n a_k \]
Find the value of the sum:
\[ \sum _{k=1}^{10} \left (2k^2+5\right ) \]
First, we’ll use the properties above to
split this into two sums, then factor the \(2\) out of the first sum.
\begin{align*} \sum _{k=1}^{10} \left (2k^2+5\right ) &= \sum _{k=1}^{10} 2k^2 + \sum _{k=1}^{10} 5\\ &= 2 \sum _{k=1}^{10} \answer [given]{k^2} + \sum _{k=1}^{10} 5 \end{align*}
The two sums we have left, can be found using formulas 1 and 3 above! We see that:
\[ \sum _{k=1}^{10} k^2 = \frac {\answer [given]{10}(\answer [given]{10}+1)(2\cdot \answer [given]{10}+1)}{6} = \answer [given]{385} \]
Similarly: \[ \sum _{k=1}^{10} 5 = \answer [given]{50} \]
Putting all that together, \(\sum _{k=1}^{10}\left (2k^2+5\right ) = 2 \cdot 385 + 50 = \answer [given]{820}\). Find the value of the sum:
\[ \sum _{k=1}^{200} \left (-6k^2+3\right ) \]
Let’s use the same approach as in the
previous example. First, we’ll use the properties to split this into
individual sums, then factor out the coefficients. After that, we’ll use the
formulas above to evaluate it. \begin{align*} \sum _{k=1}^{200} \left (-6k^2+3\right ) &= \sum _{k=1}^{200} -6k^2 + \sum _{k=1}^{200}3\\ &= - 6 \sum _{k=1}^{200} \answer [given]{k^2} + \sum _{k=1}^{200}\answer [given]{3}\\ &= - 6 \left ( \frac {200(201)(401)}{6}\right ) + 200 \cdot 3\\ &= \answer [given]{-16119600} \end{align*}
The numbers in this example were horribly ugly, but we were able to evaluate the sum without having to actually calculate all 200 terms, then add them all up. In \(4\) small lines, we were able to add \(200\) numbers.
Try one on your own.
Find the value of the sum \(\sum _{k=1}^{50} \left ( 4k^2-18k + 2(-1)^k \right )\)
2025-01-06 19:19:19 \[ \sum _{k=1}^{50} \left ( 4k^2 - 18k + 2(-1)^k \right ) = \answer {148750} \]