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Learn Gauss's method for calculating arithmetic series sums, then use the formula in worked examples.
---
## 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.
Visible text: 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 to : . 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 $$1 + 2 + 3 + \dots + 100$$.
Visible text: For example, is an arithmetic sequence with the first term and a common difference . The corresponding arithmetic series is .
### 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!
Visible text: - If the first term is added to the last term , the result is .
- If the second term is added to the second-to-last term , the result is also .
- If the third term is added to the third-to-last term , the result is still .
- This pattern continues!
Component: MathContainer
Children:
```math
\underbrace{1+100}_{101}, \underbrace{2+99}_{101}, \underbrace{3+98}_{101}, \dots, \underbrace{50+51}_{101}
```
It turns out there are $$50$$ pairs of numbers, each summing to $$101$$. So, the total sum is $$50 \times 101 = 5050$$. Clever, right?
Visible text: It turns out there are pairs of numbers, each summing to . So, the total sum is . 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 $$S_n$$.
Visible text: We can use Gauss's method to derive a general formula for the sum of the first terms of an arithmetic series, usually denoted by .
Let's say we have an arithmetic series:
```math
S_n = U_1 + U_2 + U_3 + \dots + U_{n-1} + U_n
```
If written out with the first term $$(a)$$ and the common difference $$(b)$$:
Visible text: If written out with the first term and the common difference :
```math
S_n = a + (a+b) + (a+2b) + \dots + (a+(n-2)b) + (a+(n-1)b)
```
Now, let's rewrite the series $$S_n$$ in reverse order, from the last term to the first:
Visible text: Now, let's rewrite the series in reverse order, from the last term to the first:
```math
S_n = U_n + U_{n-1} + \dots + U_2 + U_1
```
Or:
```math
S_n = (a+(n-1)b) + (a+(n-2)b) + \dots + (a+b) + a
```
Next, let's add these two versions of $$S_n$$ together, term by term:
Visible text: Next, let's add these two versions of together, term by term:
Component: MathContainer
Children:
```math
S_n = a + (a+b) + \dots + (a+(n-2)b) + (a+(n-1)b)
```
```math
S_n = (a+(n-1)b) + (a+(n-2)b) + \dots + (a+b) + a
```
```math
2S_n = [2a+(n-1)b] + [2a+(n-1)b] + \dots + [2a+(n-1)b] + [2a+(n-1)b]
```
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.
Visible text: Notice! The sum of each pair of terms (top and bottom) is always the same, which is . Since there are terms, there are such identical sums.
So, we get:
```math
2S_n = n \times (2a + (n-1)b)
```
By dividing both sides by $$2$$, we obtain the formula for the sum of the first $$n$$ terms of an arithmetic series:
Visible text: By dividing both sides by , we obtain the formula for the sum of the first terms of an arithmetic series:
```math
S_n = \frac{n}{2} (2a + (n-1)b)
```
## Practical Formulas for Arithmetic Series
There are two main formulas commonly used to calculate $$S_n$$:
Visible text: There are two main formulas commonly used to calculate :
1. If the **first term $$(a)$$** and the **common difference $$(b)$$** are known:
```math
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:
```math
S_n = \frac{n}{2}(a + a + (n - 1)b)
```
```math
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 $$(\frac{n}{2})$$.
Visible text: 1. If the **first term ** and the **common difference ** are known:
2. If the **first term ** and the **-th term ** are known:
Recall the formula for the -th term is . Substituting this into the first formula:
This second formula resembles Gauss's method: the sum of the first and last terms, multiplied by the number of pairs .
**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
Visible text: - = Sum of the first terms
- = Number of terms
- = First term ()
- = Common difference (difference between terms)
- = The
-th term
## Example Problems
### First Problem
Recalculate the sum of the series $$1 + 2 + 3 + \dots + 98 + 99 + 100$$.
Visible text: Recalculate the sum of the series .
Given:
- First term $$a = 1$$
- Last term $$U_n = U_{100} = 100$$
- Number of terms $$n = 100$$
Visible text: - First term
- Last term
- Number of terms
Since $$a$$ and $$U_n$$ are known, we use the second formula:
Visible text: Since and are known, we use the second formula:
Component: MathContainer
Children:
```math
S_n = \frac{n}{2}(a + U_n)
```
```math
S_{100} = \frac{100}{2}(1 + 100)
```
```math
S_{100} = 50(101)
```
```math
S_{100} = 5050
```
The result is exactly the same as Gauss's calculation!
### Second Problem
Given the arithmetic series: $$13 + 16 + 19 + 22 + \dots$$. Calculate the sum of the first $$30$$ terms $$(S_{30})$$!
Visible text: Given the arithmetic series: . Calculate the sum of the first terms !
Given:
- First term $$a = 13$$
- Common difference $$b = 16 - 13 = 3$$
- Number of terms to sum $$n = 30$$
Visible text: - First term
- Common difference
- Number of terms to sum
Since $$a$$ and $$b$$ are known, we use the first formula:
Visible text: Since and are known, we use the first formula:
Component: MathContainer
Children:
```math
S_n = \frac{n}{2}(2a + (n - 1)b)
```
```math
S_{30} = \frac{30}{2}[2(13) + (30 - 1)(3)]
```
```math
S_{30} = 15[26 + (29)(3)]
```
```math
S_{30} = 15(26 + 87)
```
```math
S_{30} = 15(113)
```
```math
S_{30} = 1695
```
So, the sum of the first $$30$$ terms of this series is $$1695$$.
Visible text: So, the sum of the first terms of this series is .