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(1.3) Arithmetic Series

Exploring Arithmetic Series

Find the sum of $1+2+3+\cdots+98+99+100$.

How many terms are in this sequence?

Let $S_n$ mean the sum of the first $n$ terms of a series. Because our example has 100 terms,
$S_{100}=1+2+3+…+98+99+100$

If we wrote this series in reverse order, it would look like
$S_{100}=100+99+98+…+3+2+1$

Consider adding these two equations:
$$
\begin{array}{lc}
S_{100}=&\color{red}{1}&+&2&+&3&+&…&+&98&+&99&+&100 \\
S_{100}=&\color{red}{100}&+&99&+&98&+&…&+&3&+&2&+&1 \\
\hline
\end{array}
$$

What is the sum of $\color{red}{1}$ and $\color{red}{100}$?

$$
\begin{array}{lc}
S_{100}=&1&+&\color{red}{2}&+&3&+&…&+&98&+&99&+&100 \\
S_{100}=&100&+&\color{red}{99}&+&98&+&…&+&3&+&2&+&1 \\
\hline
&101\\
\end{array}
$$

What is the sum of $\color{red}{2}$ and $\color{red}{99}$?

$$
\begin{array}{lc}
S_{100}=&1&+&2&+&\color{red}{3}&+&…&+&98&+&99&+&100 \\
S_{100}=&100&+&99&+&\color{red}{98}&+&…&+&3&+&2&+&1 \\
\hline
&101&+&101\\
\end{array}
$$

What is the sum of $\color{red}{3}$ and $\color{red}{98}$?

$$
\begin{array}{lc}
S_{100}=&1&+&2&+&3&+&…&+&98&+&99&+&100 \\
S_{100}=&100&+&99&+&98&+&…&+&3&+&2&+&1 \\
\hline
&101&+&101&+&101&+&…&+&101&+&101&+&101\\
\end{array}
$$

After summing all pairs, how many $101$'s would there be?

Therefore, the sum of all the $101$'s equals

$$
\begin{array}{rc}
S_{100}=&1&+&2&+&3&+&…&+&98&+&99&+&100 \\
S_{100}=&100&+&99&+&98&+&…&+&3&+&2&+&1 \\
\hline
\color{red}{2S_{100}}=&101&+&101&+&101&+&…&+&101&+&101&+&101\\
\color{red}{2S_{100}}=&\color{blue}{10100}
\end{array}
$$

Since $\color{blue}{10100}$ is the sum of $\color{red}{2S_{100}}$, $S_{100}=$

Sum of a Finite Arithmetic Series

Let’s find the sum of a general arithmetic series.
$$
\begin{array}{rc}
S_n=&u_1&+&\underbrace{u_1+d}_{u_2}&+&\underbrace{u_1+2d}_{u_3}&+&…&+\;\underbrace{\fbox{ 3rd last }}_{u_{n-2}}+\underbrace{\fbox{ 2nd last }}_{u_{n-1}}\;+&\underbrace{u_1+(n-1)d}_{u_n}
\end{array}
$$

Express the $\fbox{ 2nd last }$ term in terms of $u_1, n$ and $d$.
*Write your answer in the form $u_1+(n-\square)d$

Express the $\fbox{ 3rd last }$ term in terms of $u_1, n$ and $d$.
*Write your answer in the form $u_1+(n-\square)d$

Writing $S_n$ in reverse order and adding the two, we get
$$
\begin{array}{rc}
S_n=&\color{red}{u_1}&+&u_1+d&+&u_1+2d&+&…&+&u_1+(n-2)d&+&u_1+(n-1)d\\
S_n=&\color{red}{u_1+(n-1)d}&+&u_1+(n-2)d&+&u_1+(n-3)d&+&…&+&u_1+d&+&u_1\\
\hline
\end{array}
$$

What is the sum of $\color{red}{u_1}$ and $\color{red}{u_1+(n-1)d}$?
*Write your answer in the form $\square u_1+(n-\square)d$

$$
\begin{array}{rc}
S_n=&u_1&+&\color{red}{u_1+d}&+&u_1+2d&+&…&+&u_1+(n-2)d&+&u_1+(n-1)d\\
S_n=&u_1+(n-1)d&+&\color{red}{u_1+(n-2)d}&+&u_1+(n-3)d&+&…&+&u_1+d&+&u_1\\
\hline
&2u_1+(n-1)d
\end{array}
$$

What is the sum of $\color{red}{u_1+d}$ and $\color{red}{u_1+(n-2)d}$?
*Write your answer in the form $\square u_1+(n-\square)d$

$$
\begin{array}{rc}
S_n=&u_1&+&u_1+d&+&\color{red}{u_1+2d}&+&…&+&u_1+(n-2)d&+&u_1+(n-1)d\\
S_n=&u_1+(n-1)d&+&u_1+(n-2)d&+&\color{red}{u_1+(n-3)d}&+&…&+&u_1+d&+&u_1\\
\hline
&2u_1+(n-1)d&+&2u_1+(n-1)d
\end{array}
$$

What is the sum of $\color{red}{u_1+2d}$ and $\color{red}{u_1+(n-3)d}$?
*Write your answer in the form $\square u_1+(n-\square)d$

$$
\begin{array}{rc}
S_n=&u_1&+&u_1+d&+&\color{red}{u_1+2d}&+\;…\;+&u_1+(n-2)d&+&u_1+(n-1)d\\
S_n=&u_1+(n-1)d&+&u_1+(n-2)d&+&\color{red}{u_1+(n-3)d}&+\;…\;+&u_1+d&+&u_1\\
\hline
2S_n=&2u_1+(n-1)d&+&2u_1+(n-1)d&+&2u_1+(n-1)d&+\;…\;+&2u_1+(n-1)d&+&2u_1+(n-1)d
\end{array}
$$

How many $2u_1+(n-1)d$ 's are there?

Therefore, $2S_n=$

Finally, $S_n=$
Example 16

Find the sum of $4+7+10+13+…$ to $50$ terms.

If you know the last term of the sequence,
$$
\begin{array}{rc}
S_n=&\color{red}{u_1}&+&u_2&+&u_3&+&…&+&u_{n-1}&+&u_n\\
S_n=&\color{red}{u_n}&+&u_{n-1}&+&u_{n-2}&+&…&+&u_2&+&u_1\\
\hline
2S_n=&\color{red}{u_1+u_n}
\end{array}
$$

The formula can be simplified to $S_n=$
Example 17

Find the sum of $-6+1+8+15+…+141$.