Sayber and Tuscany sell popsicles during the summer for pocket money. One particular weekend, they purchased a package of 30 popsicles from the store.
Usually they just offer the popsicles for free, 1 per customer, and accept tips. This time, Tuscany wonders if they would make more money by charging $0.50 per popsicle. At the same time, Sayber wonders if he might be able to increase his tips by encouraging customers to "outbid" each other. The two children decide to each take 15 popsicles and see who makes the most.
How could you calculate how much money each of them makes, assuming Sayber gets a $0.10 tip from the first customer, and is able to convince each successive customer to double the previous person's tip?
Watch This
Guidance
Consider for example a sequence defined by a_{n} = 3n. If we write out the sum of the first 4 terms, we have 3 + 6 + 9 + 12 = 30. But what if we want to write out terms for a larger sum?
Summation notation is a method of writing sums in a succinct form. To write the sum 3 + 6 + 9 + 12 = 30 , we use the Greek letter Sigma, as follows:
\begin{align*}\sum_{n=1}^4 3n\end{align*} 

The expression 3n is called the summand, the 1 and the 4 are referred to as the limits of the summation, and the n is called the index of the sum. Here we have used a “sigma” to write a sum. We can also read a sigma, and determine the sum. For example, we can read the above sigma notation as “find the sum of the first four terms of the series, where the n^{th} term is 3n.” We always read the limits from the bottom to the top. The bottom number tells you which term to start with, and the top limit tells you which term is the final term to add. We could then write out the sigma above as:
\begin{align*}\sum_{n=1}^4 3n\end{align*} 
= 3(1) + 3(2) + 3(3) + 3(4)  

= 3 + 6 + 9 + 12 = 30 
In general, we can either rewrite a given series in sigma notation, or we can read sigma notation in order to find the value of the sum.
Properties of Sigma
Notice that we can write the sum 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 19 + 20 as 2(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10). Therefore \begin{align*}\sum_{n=1}^{10} 2n = 2 \sum_{n=1}^{10} n\end{align*}
\begin{align*}\sum_{n=1}^{k}ca_n \end{align*} 
\begin{align*} = c\sum_{n=1}^{k} a_n \end{align*} 

Example A
Write the sum using sigma notation:
 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 19 + 20
Solution:
\begin{align*}\sum_{n=1}^{10} 2n\end{align*}
Every term is a multiple of 2. The first term is 2 × 1, the second term is 2 × 2 , and so on. So the summand of the sigma is 2n. There are 10 terms in the sum. Therefore the limits of the sum are 1 and 10.
Example B
Write out the terms of \begin{align*}\sum_{n=2}^{5}(n+7) \end{align*}
Solution:
Break the sum into two different sums. The sum is 37.
\begin{align*}\sum_{n=2}^{5}(n+7) \end{align*} 
=(2 + 7) + (3 + 7) + (4 + 7) + (5 + 7)  

=2 + 3 + 4 + 5 + 7 + 7 + 7 + 7  
=2 + 3 + 4 + 5 + 7 × 4 = 14 + 28 = 42 
Notice we could have written \begin{align*}\sum_{n=2}^{5}(n+7) \end{align*}
Example C
Write out the terms of \begin{align*}\sum_{n=0}^{4}32\left ( \frac{1}{4} \right )^n\end{align*}
Solution:
In general, we can write a sum as a sum of sums: \begin{align*}\sum_{n=1}^{n}(a_n+b_n) =\sum_{n=1}^{n}(a_n)+\sum_{n=1}^{n}(b_n)\end{align*}
The sum is \begin{align*}42\frac{5}{8}\end{align*}
\begin{align*}\sum_{i=0}^{4}32 \left ( \frac{1}{4} \right)^n\end{align*} 
\begin{align*}= 32 \left ( \frac{1}{4} \right )^0+32 \left ( \frac{1}{4} \right )^1+32 \left ( \frac{1}{4} \right )^2+32 \left ( \frac{1}{4} \right )^3+32 \left ( \frac{1}{4} \right )^4\end{align*} 


\begin{align*}= 32 \cdot 1+32\cdot \frac{1}{4}+32 \cdot \frac{1}{16}+
32\cdot \frac{1}{64}+32 \cdot \frac{1}{256}\end{align*} 

\begin{align*}= 32+8+2+\frac{1}{2}+\frac{1}{8}=42 \frac{5}{8}\end{align*} 
Concept question wrapup:


>
Guided Practice
1) 2 + 6 + 18 + 54

\begin{align*}\sum_{n = 1}^4 2 \left(3^{n  1} \right)\end{align*}
∑n=142(3n−1) or \begin{align*}\sum_{n = 0}^3 2 \left(3^n \right)\end{align*}∑n=032(3n)
2) Find the series of numbers and the total of those numbers of the arithmetic series represented by the following sigma notation: \begin{align*}\sum_{n=8}^{14} {3 + \frac{3}{4}(n1)}\end{align*}
3) Find the sum of all the numbers in the arithmetic sequence. \begin{align*}\sum_{n=8}^{28} {9  2(n1)}\end{align*}
4) Find the series of numbers and the total of those numbers of the geometric series represented by the sigma notation \begin{align*}\sum_{n=1}^8 7(\frac{1}{2})^{n1}\end{align*}
5) Find the sum of the terms in: \begin{align*}\sum_{n=1}^{11} 9(4)^{n1}\end{align*}
Answers
1) \begin{align*}\sum_{n = 1}^5 2n  1\end{align*}
2) To find the series of numbers, we plug in all the numbers between 8 and 14 for (n)


a) \begin{align*} 3 + \frac{3}{4}((8)1) = \frac{33}{4}\end{align*}
3+34((8)−1)=334 
b) \begin{align*} 3 + \frac{3}{4}((9)1) = 9\end{align*}
3+34((9)−1)=9

a) \begin{align*} 3 + \frac{3}{4}((8)1) = \frac{33}{4}\end{align*}
We would continue to do this clear through the number 14. Leaving us with the following series: \begin{align*} \frac{33}{4} + 9 + \frac{39}{4} + \frac{21}{2} + \frac{45}{4} + 12 + \frac{51}{4}\end{align*}
 This is fine, if we are just looking for the individual numbers in the sequence, however when asked to evaluate sumations we are being asked to add all the numbers of the series together. It took plenty long enough to find each number, and now we must add them all together. Fortunately there is a formula, not only eliminating our need to find each number, but that lets us also add them all together and arrive at our answer or sum.
The formula is \begin{align*}\frac{k}{2}(a_1 + a_{n})\end{align*}

We plug in n = 8 to get \begin{align*}a_8 = \frac{33}{4}\end{align*}
a8=334 
Then we plug in n = 14, to get \begin{align*}a_{14} = \frac{51}{4}\end{align*}
a14=514 
Identify the number of terms \begin{align*}(k) = 14  8 + 1\end{align*}
(k)=14−8+1 so we use \begin{align*}k=7\end{align*}k=7 in the formula below. 
Now we can use the formula:\begin{align*}\frac{k}{2}(a_8 + a_{14})\end{align*}
k2(a8+a14) and we get:\begin{align*} \frac{7}{2}(\frac{33}{4} + \frac{51}{4})=\frac{147}{2}\end{align*}72(334+514)=1472
\begin{align*}\therefore \sum_{n = 1}^4 2 \left(3^{n  1} \right)\end{align*} or \begin{align*}\sum_{n = 0}^3 2 \left(3^n \right)\end{align*}
3) Use the formula \begin{align*}\frac{k}{2}(a_1 + a_{n})\end{align*} to find the sum of \begin{align*}\sum_{n=8}^{28} {9  2(n1)}\end{align*}
 Substitute n = 8 to get \begin{align*}a_8 = 5\end{align*}
 Substitute n = 28 to get \begin{align*}a_28 = 45\end{align*}
 Identify the number of terms as: \begin{align*}k = 288+1\end{align*} so we use \begin{align*}k = 21\end{align*} in our formula below:
 Now we can use the formula:\begin{align*}\frac{k}{2}(a_8 + a_{14})\end{align*} and our answer is \begin{align*}525\end{align*}
\begin{align*}\therefore \sum_{n = 1}^{10} \left(\frac{1} {n} \right)\end{align*}
4) To identify and sum the terms in the geometric series \begin{align*}\sum_{n=1}^8 7(\frac{1}{2})^{n1}\end{align*}
 a) Find the sequence of numbers the same way as we did in problem 2, by plugging in the indicated numbers 18 for (n)
 b) \begin{align*}\sum_{n=1}^8 7(\frac{1}{2})^{((1)1)}\end{align*} which gives us: \begin{align*} 7 \end{align*}
 c) We do this for the remaining numbers and our sequence looks like this:\begin{align*}7 + \frac{7}{2} +\frac{7}{4} + \frac{7}{8} +\frac{7}{16} +\frac{7}{32} +\frac{7}{64} + \frac{7}{128}\end{align*}
As in prior examples, we are not just being asked to find the numbers, but add them all together. Again, we do not want to have to take the time to find all the numbers and add them all together. Once again we find ourselves lucky, there is a formula!
The formula \begin{align*}a_1(\frac{1r^k}{1r})\end{align*} works like this:
 We plug n=1 into \begin{align*}a_1\end{align*} which gives us 7.
 \begin{align*}k = 8\end{align*}, and \begin{align*}r=\frac{1}{2}\end{align*}
 Substituting our numbers results:\begin{align*}7(\frac{1\frac{1}{2}^8}{1\frac{1}{2}})\end{align*}
 Which gives us \begin{align*}\frac{595}{128}\end{align*}
5) Let's use the formula for geometric series again:
 Identify the terms: \begin{align*}a_1 = 9\end{align*}, \begin{align*} k = 11\end{align*} and \begin{align*} r = 4\end{align*}
 Substituting into our formula, we have: \begin{align*}9(\frac{14^{11}}{14}) = 12,582,909\end{align*}
Explore More
Express the sum using sigma notation:
 \begin{align*}1 + 3 + 5 + 7 + 9\end{align*}
 \begin{align*}1 + \frac{1} {2} + \frac{1} {3} + \frac{1} {4} + ... + \frac{1} {10}\end{align*}
Find the series of numbers indicated by the sigma notation:
 \begin{align*}\sum_{n=0}^{14} 2 \frac{10}{3}(n1)\end{align*}
 \begin{align*}\sum_{n=8}^{14} 7 3(n1)\end{align*}
Evaluate:
 \begin{align*}\sum_{n=10}^{5} 7 \frac{4}{3}(n1)\end{align*}
 \begin{align*}\sum_{n=3}^{3} 3 \frac{1}{3}(n1)\end{align*}
 \begin{align*}\sum_{n=5}^{1} 6 +\frac{4}{3}(n1)\end{align*}
Find the series of numbers indicated by the sigma notation:
 \begin{align*}\sum_{n=1}^{2} 3(\frac{1}{2})^{n1}\end{align*}
 \begin{align*}\sum_{n=1}^{5} 5(\frac{4}{3})^{n1}\end{align*}
Evaluate:
 \begin{align*}\sum_{n=1}^{6} 3(\frac{1}{2})^{n1}\end{align*}
 \begin{align*}\sum_{n=1}^{7} 5(\frac{1}{2})^{n1}\end{align*}
 \begin{align*}\sum_{n=1}^{11} 7(\frac{4}{3})^{n1}\end{align*}
 \begin{align*}\sum_{n=1}^{6} 7(\frac{1}{4})^{n1}\end{align*}
 \begin{align*}\sum_{n=1}^{3} 2(\frac{3}{2})^{n1}\end{align*}