The sum of terms in a sequence is called a series. The sum of the first $n$n terms of a sequence is denoted $S_n$Sn, that is $S_n=t_1+t_2+t_3+...t_n$Sn=t1+t2+t3+...tn. Let's try and find a formula for calculating the first $n$n terms of an arithmetic sequence.
What would be the sum of the first $100$100 integers? An oft quoted anecdote has the famous mathematician Carl Friedrich Gauss, at the age of seven, set the task by his teacher to add up the first $100$100 integers. The teacher is said to have set the task to keep Gauss busy for a while but he replied almost immediately with the answer. Let's look at the way this was calculated and see if we can generalise this to any arithmetic sequence.
First write down the sum in order:
$S_{100}=1+2+3+4+5+.......+96+97+98+99+100$S100=1+2+3+4+5+.......+96+97+98+99+100
Then underneath write the sum in reverse order:
| $S_{100}=$S100= |
$1$1 |
$+2$+2 | $+3$+3 | $+4$+4 | $...$... | $+97$+97 | $+98$+98 | $+99$+99 | $+100$+100 |
| $S_{100}=$S100= | $100$100 | $+99$+99 | $+98$+98 | $+97$+97 | $...$... | $+4$+4 | $+3$+3 | $+2$+2 | $+1$+1 |
Adding term-wise vertically we get the sum:
$2S_{100}=101+101+101+101+..........+101+101+101+101$2S100=101+101+101+101+..........+101+101+101+101
Since we added $100$100 terms we have:
$2S_{100}=100(101)$2S100=100(101)
And hence, $S_{100}=\frac{100}{2}(101)=5050$S100=1002(101)=5050.
So the first $100$100 integers add to $5050$5050. This was an arithmetic sequence with $t_1=1$t1=1 and $d=1$d=1. Let's we try this process on a general arithmetic sequence of length $n$n with starting value $t_1$t1 and common difference $d$d.
$S_n=t_1+\left(t_1+d\right)+\left(t_1+2d\right)+....+\left(t_1+\left(n-3\right)d\right)+\left(t_1+\left(n-2\right)d\right)+\left(t_1+\left(n-1\right)d\right)$Sn=t1+(t1+d)+(t1+2d)+....+(t1+(n−3)d)+(t1+(n−2)d)+(t1+(n−1)d)
Then writing this sum again underneath but in reverse order, we get:
Adding term-wise vertically each pair of terms adds to $2t_1+\left(n-1\right)d$2t1+(n−1)d and we have $n$n of these terms so:
$2S_n=n\left(2t_1+\left(n-1\right)d\right)$2Sn=n(2t1+(n−1)d)
And hence, the sum of the first $n$n terms of an arithmetic sequence is given by:
$S_n=\frac{n}{2}\left(2t_1+\left(n-1\right)d\right)$Sn=n2(2t1+(n−1)d)
The $n$nth term of an arithmetic sequence is: $t_n=t_1+\left(n-1\right)d$tn=t1+(n−1)d, if we are finding the sum of the first $n$n terms then is will be our last term, $l$l. If we know the last term, as we did when adding the first $100$100 integers, we can simplify the formula to:
$S_n=\frac{n}{2}\left(t_1+l\right)$Sn=n2(t1+l)
For any arithmetic sequence with starting value $t_1$t1 and common difference $d$d, we can find the sum of the first $n$n terms, using:
$S_n=\frac{n}{2}\left(2t_1+\left(n-1\right)d\right)$Sn=n2(2t1+(n−1)d)
OR
$S_n=\frac{n}{2}\left(t_1+l\right)$Sn=n2(t1+l), where $l$l is the last term of the sequence$\left(l=t_n=t_1+\left(n-1\right)d\right)$(l=tn=t1+(n−1)d)
Worked examples
Example 1
Find the sum of the multiples of $7$7 between $100$100 and $200$200.
Think: What is the first multiple of seven greater than $100$100, what is the last multiple of seven before $200$200 and how many multiples are there?
Do: By dividing the numbers $101,102,103,...$101,102,103,... by seven we find the first multiple above $100$100 is $t_1=105$t1=105 and similarly the last multiple before $200$200 is $l=196$l=196. The number of multiples, $n$n, can be found from dividing the difference between our first and last value by $7$7 and adding $1$1, to count our first multiple. Hence, $n=\frac{196-105}{7}+1=14$n=196−1057+1=14. Alternatively, this can be found by considering we have the sequence $105,112,119...196$105,112,119...196, with $t_1=105$t1=105, $d=7$d=7 and solving $t_n=196$tn=196 for $n$n using $t_n=t_1+(n-1)d$tn=t1+(n−1)d.
For the summation, using $S_n=\frac{n}{2}\left(t_1+l\right)$Sn=n2(t1+l), we have:
| $S_{14}$S14 | $=$= | $\frac{14}{2}\left(105+196\right)$142(105+196) |
| $=$= | $2100$2100 |
Example 2
Rows of stands to take school photos are set up with $2$2 additional places in each successive row.
a) If there are $10$10 students in the first row how many students in total would there be in $5$5 rows?
Think: We want to sum of the first $5$5 terms of an arithmetic sequence that starts with $10$10 and the common difference is $2$2. Write down $n$n, $t_1$t1 and $d$d and use the formula $S_n=\frac{n}{2}\left(2t_1+\left(n-1\right)d\right)$Sn=n2(2t1+(n−1)d) to find the sum.
Do: $n=5$n=5, $t_1=10$t1=10 and $d=2$d=2, substituting into the formula we get:
| $S_5$S5 | $=$= | $\frac{5}{2}\left(2\times10+\left(5-1\right)2\right)$52(2×10+(5−1)2) |
| $=$= | $\frac{5}{2}\left(20+8\right)$52(20+8) | |
| $=$= | $70$70 |
There would be $70$70 students in the photo.
b) If there were $15$15 students in the first row, how many rows are required for a shot of a year level with $159$159 students.
Think: We have the sum but require $n$n. We need to write down $t_1$t1, $d$d, and $S_n$Sn, substitute these values into $S_n=\frac{n}{2}\left(2t_1+\left(n-1\right)d\right)$Sn=n2(2t1+(n−1)d), and solve for $n$n. Note: Since $n$n appears in two factors of this formula we will obtain a quadratic equation. We can then solve this by factorisation, the quadratic formula or an appropriate method using technology.
Do: $t_1=15,d=2,S_n=147$t1=15,d=2,Sn=147 and hence:
| $S_n$Sn | $=$= | $\frac{n}{2}\left(2t_1+\left(n-1\right)d\right)$n2(2t1+(n−1)d) | |
| $147$147 | $=$= | $\frac{n}{2}\left(2\times15+\left(n-1\right)2\right)$n2(2×15+(n−1)2) | Substitute in values |
| $294$294 | $=$= | $n\left(28+2n\right)$n(28+2n) | Multiply both sides by 2 and simplify |
| $294$294 | $=$= | $28n+2n^2$28n+2n2 | Expand brackets |
| $\therefore2n^2+28n-294$∴2n2+28n−294 | $=$= | $0$0 | Bring all terms to one side |
| $n^2+14n-147$n2+14n−147 | $=$= | $0$0 | |
| $(n-7)(n+21)$(n−7)(n+21) | $=$= | $0$0 | Factorise |
Hence, $n=7$n=7 as $n>0$n>0. So they will require $7$7 rows for the photo.
Practice questions
QUESTION 1
Find the sum of the first $10$10 terms of the arithmetic sequence defined by $T_1=8$T1=8 and $d=3$d=3.
QUESTION 2
Consider the arithmetic sequence $4$4, $-1$−1, $-6$−6, …
Write a simplified expression for the sum of the first $n$n terms.
Find the sum of the progression from the $18$18th to the $27$27th term, inclusive.
QUESTION 3
A worker at a factory is stacking cylindrical-shaped pipes which are stacked in layers. Each layer contains one pipe less than the layer below it. There are $6$6 pipes in the topmost layer, $7$7 pipes in the next layer, and so on. There are $n$n layers in the stack.
Form an expression for the number of pipes in the bottom layer.
Show that there are a total of $n\left(\frac{n+11}{2}\right)$n(n+112) pipes in the stack.
QUESTION 4
Quentin is learning to drive. His first lesson is $22$22 minutes long, and each subsequent lesson is $3$3 minutes longer than the lesson before.
How long will his $15$15th lesson be?
If Quentin reaches $22.6$22.6 total hours of driving on his $n$nth lesson, solve for $n$n.
Summation notation
If we were calculating $2+4+6+8+10.....+30$2+4+6+8+10.....+30, we can write this in a compact notation as:
$\sum_{n=1}^{15}2n$15∑n=12n
The symbol $\Sigma$Σ (pronounced "sigma") is the capital letter S in the Greek alphabet. When $\Sigma$Σ is used to express a series in mathematics, then in those instances it stands for the word "Sum".
So the expression above can be read as "the sum of $2n$2n for the values $n=1$n=1 to $15$15".
Worked examples
Example 3
Evaluate the expression:
$\sum_{n=1}^5$5∑n=1 $n^2$n2
The equation $n=1$n=1 directly below the $\Sigma$Σ sign tells us that we start the series by substituting $n=1$n=1 into the formula $n^2$n2. So the series begins:
$1^2$12
Then we increase $n$n by $1$1, so that $n$n becomes $2$2. The new value of $n$n is substituted into the sequence formula to reveal $2^2$22 so that the series becomes:
$1^2+2^2$12+22
Again we increase $n$n by $1$1 and substitute for the third term so that the series becomes:
$1^2+2^2+3^2$12+22+32
This process of increasing $n$n by $1$1 continues until $n=5$n=5 is reached as shown directly above the $\Sigma$Σ sign. The complete sum becomes:
$\sum_{n=1}^5n^2=1^2+2^2+3^2+4^2+5^2$5∑n=1n2=12+22+32+42+52
Thus this series sums to $55$55.
Example 4
Write the sum of the first $100$100 terms of the arithmetic series whose first term is $t_1=10$t1=10 and whose common difference is $d=3$d=3, with summation notation and evaluate the expression.
Think: We want to sum from term $1$1 to term $100$100 for the sequence with the rule $t_n=10+(n-1)\times3$tn=10+(n−1)×3, which is equivalent to $t_n=7+3n$tn=7+3n.
We could write this as: $\sum_{n=1}^{100}t_n$100∑n=1tn where $t_n=7+3n$tn=7+3n. Or we can write it is a single expression as:
$\sum_{n=1}^{100}\left(7+3n\right)$100∑n=1(7+3n)
Reflect: Substitute for $n=1,2,3...$n=1,2,3... the first few terms to see how the formula unfolds:
$10+13+16+...+304+307$10+13+16+...+304+307
To evaluate the summation we can use our formula: $S_n=\frac{n}{2}\left(t_1+l\right)$Sn=n2(t1+l):
| $S_{100}$S100 | $=$= | $\frac{100}{2}\left(10+307\right)$1002(10+307) |
| $=$= | $15850$15850 |
Summation notation is quite versatile in its application to series. For example, the mean of a set of scores, say in general terms the scores $x_1,x_2,x_3,...,x_n$x1,x2,x3,...,xn is their sum divided by $n$n.
We can write this using $i$i to indicate the score number and write:
$\sum_{i=1}^n$n∑i=1 $\frac{x_i}{n}$xin
In expanded form this expression reads:
$\frac{x_1}{n}+\frac{x_2}{n}+\frac{x_3}{n}+...+\frac{x_n}{n}$x1n+x2n+x3n+...+xnn
or equivalently $\frac{x_1+x_2+x_3+...+x_n}{n}$x1+x2+x3+...+xnn.
Practice questions
Question 5
Write the following series using sigma notation.
$4+8+12+16+20+\text{. . .}$4+8+12+16+20+. . .
$\sum_{k=1}^{\infty}\left(\editable{}\right)$∞∑k=1()
Question 6
Consider the series:
$\frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\frac{1}{5^3}$113+123+133+143+153
Rewrite the series using sigma notation in the form $\sum_{k=\editable{}}^{\editable{}}\editable{}$∑k=.
Question 7
Find the value of $\sum_{r=1}^4\frac{1}{r+2}$4∑r=11r+2.