Arithmetic Series

Understanding Arithmetic Series

Ever heard the story about Carl Friedrich Gauss, the math genius? When he was in elementary school, his teacher assigned the task of summing all numbers from 1 to 100: $1 + 2 + 3 + \dots + 98 + 99 + 100$. The teacher hoped this would keep the students busy for a while.

But Gauss had a brilliant idea! He didn't sum them one by one. This sequential summation of terms from an arithmetic sequence (a sequence with a constant difference between terms) is what we call an Arithmetic Series.

For example, $1, 2, 3, \dots, 100$ is an arithmetic sequence with the first term $a=1$ and a common difference $b=1$. The corresponding arithmetic series is $1 + 2 + 3 + \dots + 100$.

How Did Gauss Calculate It?

Gauss noticed an interesting pattern:

• If the first term $(1)$ is added to the last term $(100)$, the result is $101$.
• If the second term $(2)$ is added to the second-to-last term $(99)$, the result is also $101$.
• If the third term $(3)$ is added to the third-to-last term $(98)$, the result is still $101$.
• This pattern continues!

It turns out there are 50 pairs of numbers, each summing to 101. So, the total sum is $50 \times 101 = 5050$. Clever, right?

Finding the General Formula

We can use Gauss's method to derive a general formula for the sum of the first $n$ terms of an arithmetic series, usually denoted by $S_n$.

Let's say we have an arithmetic series:

If written out with the first term $(a)$ and the common difference $(b)$:

Now, let's rewrite the series $S_n$ in reverse order, from the last term to the first:

Or:

Next, let's add these two versions of $S_n$ together, term by term:

Notice! The sum of each pair of terms (top and bottom) is always the same, which is $2a + (n-1)b$. Since there are $n$ terms, there are $n$ such identical sums.

So, we get:

By dividing both sides by 2, we obtain the formula for the sum of the first $n$ terms of an arithmetic series:

Practical Formulas for Arithmetic Series

There are two main formulas commonly used to calculate $S_n$:

1. If the first term $(a)$ and the common difference $(b)$ are known:

   $$S_n = \frac{n}{2}(2a + (n - 1)b)$$

2. If the first term $(a)$ and the $(n)$-th term $(U_n)$ are known: Recall the formula for the $n$-th term is $U_n = a + (n-1)b$. Substituting this into the first formula:

   $$S_n = \frac{n}{2}(a + a + (n - 1)b)$$

   $$S_n = \frac{n}{2}(a + U_n)$$

   This second formula resembles Gauss's method: the sum of the first and last terms, multiplied by the number of pairs $(\frac{n}{2})$.

Notation:

• $S_n$ = Sum of the first $n$ terms
• $n$ = Number of terms
• $a$ = First term ($U_1$)
• $b$ = Common difference (difference between terms)
• $U_n$ = The $n$ -th term

Example Problems

Problem 1

Recalculate the sum of the series $1 + 2 + 3 + \dots + 98 + 99 + 100$.

Given:

• First term $a = 1$
• Last term $U_n = U_{100} = 100$
• Number of terms $n = 100$

Since $a$ and $U_n$ are known, we use the second formula:

The result is exactly the same as Gauss's calculation!

Problem 2

Given the arithmetic series: $13 + 16 + 19 + 22 + \dots$. Calculate the sum of the first 30 terms $(S_{30})$!

Given:

• First term $a = 13$
• Common difference $b = 16 - 13 = 3$
• Number of terms to sum $n = 30$

Since $a$ and $b$ are known, we use the first formula:

So, the sum of the first 30 terms of this series is 1695.