Arithmetic Sequences and Sums
Arithmetic Sequences
In other words, a sequence is arithmetic if you can go from one entry to the next simply by repeatedly adding the same constant number.
For example, the sequence \(A=(3,5,7,9,11,13,15,17,…)\) is arithmetic because, in order to go from \(3\) to \(5\) or from \(13\) to \(15\), or any entry \(a_i\) to the entry right after it, \(a_{i+1}\), all one has to do is add \(2\) to the previous entry to get the next entry. In this case, the constant referred to in the above definition is \(2\). Note that if you subtract any pair of consecutive entries of this sequence, you will get this constant \(2\).
Given the above observation, finding a closed formula for an arithmetic sequence is really easy!
Note that in \(A\) we can find each element of \(A\) by adding the common difference \(c\) to the previous element. So if we start from the very first element \(a_0\), we can find \(a_1\) by adding \(c\) to \(a_0\) once. To get from \(a_1\) to \(a_2\), we yet again add \(c\) to \(a_1\), which is the same as adding \(c\) to \(a_0\) twice. And so on. Thus, in the formula \(a_i=a_0+c\cdot i\), the \(i\) indicates how many times we are adding the common difference \(c\) to the starting element \(a_0\) to get to the \(i\)th element beyond it.
The common difference can be found by subtracting any pair of consecutive entries of the sequence. The common difference is therefore \(c=3\).
The closed formula for this sequence is thus \(b_i=a_0+c\cdot i= 2+3\cdot i\).
To see that this is correct, notice
\(b_0=2+3(0)=2\)
\(b_1=2+3(1)=5\)
\(b_2=2+3(2)=8\)
\(b_3=2+3(3)=11\) etc.
The entries generated from the rule match those of the original given sequence.
Finding Sums of (Terminating) Arithmetic Sequences
Suppose, like mathematician Carl Friedrich Gauss in elementary school, your teacher wants to keep you busy, and asks you to add up the first 100 natural numbers \(1,2,3,4,5,…,98,99,100\). How could one do this quickly?
Gauss (yes, in elementary school) noticed that one can pair \(1\) and \(100\), \(2\) and \(99\), \(3\) and \(98\), and so on, adding each pair to get \(101\) every time. He then noticed that there were \(50\) such pairs (100 numbers divided into groups of 2), and therefore \(50\) sums of \(101\). In total, once you add up all 50 of these \(101\)’s, you end up with a total of \(5050\).
Suppose \(S\) is the sum of the first \(100\) natural numbers. One can visualize Gauss’s process as follows:
$$\begin{align} S&=1+2+3+4+…+97+98+99+100\\+S&=100+99+98+97+…+4+3+2+1\\—&——————————————————-\\2S&=101+101+101+101+…+101+101+101+101\end{align}$$
The right hand side of the above is \(100\cdot 101=10100\) because we were adding up 100 numbers in both rows prior. This gives us the equation
$$2S=10100$$
Now, since the goal was to find the sum \(S\), and since on the left-hand-side (LHS) we have \(2S\), we can find \(S\) by dividing both sides by \(2\); getting \(S\) by itself on the LHS. This gives us
$$S=\frac{10100}{2}=5050$$
which is precisely what we were after.
Generalizing the Technique
The above pairing technique can be used to find the sum of all entries of any (terminating) arithmetic sequence.
Start by letting \(S\) be the sum of all entries of the sequence, and write the sum in reverse on the line below that. Viz:
$$\begin{align}S&=2+4+6+8+…+118+120+122\\+S&=122+120+118+…+8+6+4+2\\—-&—————————————————–\\2S&=124+124+124+124+…+124+124+124+124\end{align}$$
Where the last line comes from adding straight down the columns.
Note that to correctly compute the right-hand-side (RHS) we need to know how many terms we are adding together. The number of \(124\)’s being added together corresponds to the number of entries in the original sequence (can you see why?).
Note that the rule for sequence is given by \(a_i=a_0+c\cdot i\) where \(a_0\) is the first entry, and \(c\) is the common difference. Therefore \(a_i=2+2i\).
We want to determine which sequence element the last entry \(122\) is; i.e. how many elements we had to “go through” before we get to \(122\). Thus, we want to find \(i\) such that \(a_i=2+2i=122\) because the number of times we add \(2\) to the starting value of \(2\) corresponds to the number of elements between \(2\) and \(122\).
Solving for \(i\) in the equation \(2+2i=122\) gives you \(i=60\). Therefore, \(122\) is the 60th element of the sequence, and since it is the last element of the sequence, there must be 60 elements in the sequence. Thus, returning to our sum, we have
$$\begin{align}2S&=124+124+124+124+…+124+124+124+124\\&=60\cdot(124)\\&=7440\end{align}$$
Dividing both sides by \(2\) gets \(S\) by itself, giving you \(S=3720\), which is therefore equal to the sum of \(2,4,6,…,120,122\), as was to be found.
Common Difference: \(c=6\)
Closed Formula: \(a_i=5+6\cdot i\)
Since the sequence was assumed to be arithmetic, the common difference can be calculated by subtracting any element from the element just after it (i.e. \(a_{i+1}-a_i\)). The difference \(11-5\) found by subtracting the first and second element will suffice.
To get the closed formula, note that the sequence starts at \(5\), and that we can get from one element to the next by repeatedly adding the common difference of \(6\). To get from element \(0\) to element \(2\), say, this would require adding \(6\) to the start value \(5\) twice; to get from the start value of \(5\) to the \(i\)th value, that would require adding \(6\) \(i\) times; hence the formula you see above.
Common Difference: \(c=18\)
Closed Formula: \(a_i=2+18\cdot i\)
Since the sequence was assumed to be arithmetic, the common difference can be calculated by subtracting any element from the element just after it (i.e. \(a_{i+1}-a_i\)). The difference \(20-2\) found by subtracting the first and second element will suffice.
To get the closed formula, note that the sequence starts at \(2\), and that we can get from one element to the next by repeatedly adding the common difference of \(18\). To get from element in spot \(0\) to element in spot \(2\), say, this would require adding \(18\) to the start value \(2\) twice; to get from the start value of \(2\) to the \(i\)th value, that would require adding \(18\) \(i\) times; hence the formula you see above.
Answer: \(315\)
Answer: \(3828\)
Answer: \(4511\)