The sum of #n# terms of an arithmetic sequence can be written as:
\[s_n=t_1+t_2+t_3+\cdots+t_{n-1}+t_n=\sum_{k=1}^{n} t_k\]
If we want to calculate the sum for large values of #n#, it can be a lot of work to calculate each term and add them. Using the general formula for the sum of the terms instead is way faster.
The sum \(s_n =\sum_{k=1}^{n} t_k\) of the first #n# terms of an arithmetic sequence #t# with initial term #t_1# and difference #d# is equal to
\[s_n = n\cdot t_1+\dfrac{(n-1)\cdot n}{2}\cdot d\]
A different expression, using the last term instead of the difference, is:
\[s_n =\dfrac{n}{2} \cdot (t_1+t_n)\]
To reduce this formula, we first prove the second formula
\[s_n =\sum_{k=1}^{n} t_k=\dfrac{n}{2} \cdot (t_1+t_n)\]
Here we use an important property of arithmetic sequences, which is the fact that the sum of the first and last term is equal to the second and second-to-last term and also equal to the third and third-to-last term, etcetera. See the following:
\[t_1+t_n=t_1+t_1+(n-1) \cdot d=t_1+d+t_1+(n-2) \cdot d=t_2+t_{n-1}\]
\[\begin{array}{rclclcl}t_2+t_{n-1}&=&t_1+d+t_n- d&=&&&t_1+t_{n}\\ t_3+t_{n-2}&=&t_2+d+t_{n-1}- d&=&t_2+t_{n-1}&=&t_1+t_n\\ t_4+t_{n-3}&=&t_3+d+t_{n-2}- d&=&t_3+t_{n-2}&=&t_1+t_n\\ \vdots &=&\vdots&=&\vdots &=&t_1+t_n \end{array}\]
This means we can calculate #s_n# in the following manner:
\[\begin{array}{rcl}
t_1+t_2+t_3+\cdots +t_{n-1}+t_n&=&(t_1+t_n)+(t_2+t_{n-1})+(t_3+t_{n-2})+\cdots \\
&& \phantom{xxxxx}\color{blue}{\text{terms sorted in equal pairs}}\\
&=& (t_1+t_n)+(t_1+t_n)+(t_1+t_n)+\cdots \end{array}\]
We will check what the end of this sorting will look like. If #n# is even, then the last term is
\[ t_{\frac{n}{2}}+t_{\frac{n}{2}+1} = \left(t_{1}+t_{n}\right)\]
Hence, there are #\frac{n}{2}# terms #\left(t_{1}+t_{n}\right)#, in such a way that the sum is equal to \(s_n = \frac{n}{2}\cdot\left(t_{1}+t_{n}\right)\) like we claimed.
If #n# is odd, then the last term would have been \(\left(t_{\frac{n+1}{2}}+t_{\frac{n+1}{2}}\right)\), if it weren't for the fact that #t_{\frac{n+1}{2}}# is counted twice. The sum #s_n# hence has#\frac{n-1}{2}# terms #\left(t_{1}+t_{n}\right)# and one term \(\frac{1}{2}\left(t_{1}+t_{n}\right)\), such that
\[s_n = \left(\frac{n-1}{2}+\frac{1}{2}\right)\cdot \left(t_{1}+t_{n}\right)=\frac{n}{2}\cdot\left(t_{1}+t_{n}\right) \] like we claimed. It appears that the second formula holds for all natural numbers #n#.
To reduce the first formula from the statement, we replace #t_n# with the direct formula by #t_1+(n-1)\cdot d# in the second formula:
\[\begin{array}{rcl} s_n &=&\dfrac{n}{2} \cdot (t_1+t_n)\\ &&\phantom{xx}\color{blue}{\text{formula above}}\\ &=&\dfrac{n}{2} \cdot (t_1+t_1+(n-1)\cdot d)\\ &&\phantom{xx}\color{blue}{t_n=t_1+(n-1)\cdot d\text{ entered}}\\ &=& n\cdot t_1+\dfrac{n}{2}(n-1)\cdot d\\ &&\phantom{xx}\color{blue}{\text{brackets expanded and simplified}} \end{array}\]
The first formula shows that the sum is a quadratic function of #n#:
\[s_n = \frac{d}{2}\cdot n^2 +\left(t_1-\frac{d}{2}\right)\cdot n\]
It could happen that we want to know the sum of an infinite sequence. We call this an infinite series, which can be denoted by \[s_\infty = \sum_{k=1}^{\infty} t_k = \lim_{n\to\infty} \sum_{k=1}^{n} t_k\] For example, let #t_k=k# for #k\geq 1#. Then #t_1=1,\, t_2=2,\, t_3=3,\,\dots# The terms of this sequence get larger when #k# increases, which means the infinite series corresponding to this sequence is equal to infinity. In other words, \[ s_\infty=\sum_{k=1}^{\infty} k=\infty\] In fact, when #d# is positive, the terms of the sequence are constantly increasing in #k#, which causes #s_\infty# to be equal to infinity. On the other hand, when #d# is negative, the terms of the arithmetic sequence are constantly decreasing in #k#, which causes #s_\infty# to be equal to minus infinity. Since all infinite arithmetic series are either plus or minus infinity, we say that all infinite arithmetic series are divergent.
If an infinite series would be a real number, we would call this infinite series convergent. Later you will see examples of convergent infinite series.
With these formulas, we can calculate the sum of #n# terms of the arithmetic sequence if we know the number of terms, the first term, and the difference or the last term. Here are some examples.
What is the sum of the first #9# terms of the arithmetic sequences with #t_1=6# and #d=6#?
#\displaystyle\sum_{k=1}^{9}t_k=# #270#
To calculate the sum #\sum_{k=1}^{9}t_k# of the first #9# terms, we use the formula
\[\sum_{k=1}^{9}t_k = 9\cdot t_1+ \frac{1}{2}\cdot (9-1)\cdot 9\cdot d\]
We know that #t_1=6# and #d=6#, such that
\[\sum_{k=1}^{9}t_k = 9\cdot 6+ \frac{1}{2}\cdot 8 \cdot 9\cdot 6=270\]
Of course, we can also use the second formula from the statement. For that, we need to know #t_1# and #t_{9}#. #t_1# is given and equal to #6#. To calculate #t_{9}#, we compose the direct formula. This one is equal to: \[t_k=t_1+d \cdot (k-1)=6+6 \cdot (k-1)\]
Hence, \[t_{9}=6+6 \cdot (9-1)=54\]
Now we can use the second formula for the sum. It is:\[\sum_{k=1}^{n}t_k=\frac{1}{2} \cdot n \cdot (t_1+t_{n})\]
Hence for #9# terms this gives us: \[\sum_{k=1}^{9}t_k=\frac{1}{2} \cdot 9 \cdot (t_1+t_{9}) = \frac{1}{2} \cdot 9 \cdot (6+54)=270\]
Arithmetic series
