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Chapter 6: Problem 1

Evaluate the arithmetic series. $$ 1+2+3+\cdots+98+99+100 $$

Short Answer

Expert verified
The sum of the arithmetic series from 1 to 100 is 5050, calculated using the formula \(S_n = \frac{n(a_1 + a_n)}{2}\), with \(n = 100\), \(a_1 = 1\), and \(a_n = 100\).

Step by step solution

01

Write down the values of n, a_1, and a_n

We are given the arithmetic series: $$ 1+2+3+\cdots+98+99+100 $$ The number of terms \(n = 100\), the first term \(a_1 = 1\), and the last term \(a_n = 100\).
02

Apply the arithmetic series formula

To evaluate the sum of the arithmetic series, we use the formula for the sum of an arithmetic series: $$ S_n = \frac{n(a_1 + a_n)}{2} $$ Now, substitute the values of n, a_1, and a_n in the formula: $$ S_{100} = \frac{100(1 + 100)}{2} $$
03

Evaluate the sum

With the values substituted, let's now evaluate the sum of the series: $$ S_{100} = \frac{100(101)}{2} $$ $$ S_{100} = \frac{10100}{2} $$ $$ S_{100} = 5050 $$ So, the sum of the arithmetic series from 1 to 100 is 5050.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

arithmetic sequence
An arithmetic sequence is a list of numbers where each term after the first is obtained by adding a constant value, known as the common difference, to the previous term. In simpler words, it鈥檚 a sequence where the difference between consecutive terms is always the same. This can be thought of as a pattern of evenly spaced numbers.
For example, in the sequence \(1, 2, 3, \ldots, 100\), the common difference is \(1\). Each number increases by one as we move from one term to the next. Here:
  • First term \(a_1 = 1\)
  • Common difference \(d = 1\)
Understanding this sequence is crucial for solving problems that require finding the sum of terms or any specific term within the sequence.
sum of series
The sum of an arithmetic series involves adding all the terms of an arithmetic sequence. In this exercise, we are dealing with an arithmetic series from 1 to 100.

The key challenge is to find an efficient way to calculate this sum without manually adding each individual term. Instead of adding 1 + 2 + 3 + ... + 100 directly, there's a mathematical formula that simplifies the process.
Understanding the approach of summing these numbers will save time and confirm accuracy, especially as the number of terms increases. This understanding will also help solve similar problems involving arithmetic sequences.
arithmetic series formula
The arithmetic series formula is a straightforward equation that lets you calculate the sum of an entire series quickly. It usually takes the form:
\[ S_n = \frac{n(a_1 + a_n)}{2} \]
This formula is efficient because it uses just three key pieces of information:
  • \(n\): the total number of terms
  • \(a_1\): the first term
  • \(a_n\): the last term
In our exercise, these values were \(n = 100\), \(a_1 = 1\), and \(a_n = 100\). By substituting these into the formula, we could directly find the sum by simplifying to \( S_{100} = \frac{100(1 + 100)}{2} = 5050 \). This method showcases the power of the formula in simplifying what could be a lengthy calculation.
mathematical series evaluation
Evaluating a mathematical series is a vital skill in mathematics as it allows us to find sums quickly and efficiently. In the context of arithmetic series, the evaluation involves applying the arithmetic series formula and ensuring correct input of the sequence's parameters.
To evaluate the series \(1 + 2 + 3 + ... + 100\), we followed several key steps:
  • Recognizing it as an arithmetic sequence with a known first and last term
  • Counting the total terms \(n\)
  • Substituting these values into the arithmetic series formula
This example demonstrates how evaluation not only saves time but also ensures precision in finding the sum of a series, a critical aspect of larger mathematical and practical applications.

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Most popular questions from this chapter

Evaluate \(\lim _{n \rightarrow \infty}\left(1+\frac{3}{n}\right)^{n}\)

Evaluate \(\lim _{n \rightarrow \infty} \frac{2 n^{2}+5 n+1}{5 n^{2}-6 n+3}\)

Show that $$ \sum_{k=0}^{n} \frac{n !}{k !(n-k) !}=2^{n} $$ for every positive integer \(n\). [Hint: Expand \((1+1)^{n}\) using the Binomial Theorem.]

Show that $$ \ln n<1+\frac{1}{2}+\cdots+\frac{1}{n-1} $$ for every integer \(n \geq 2\). [Hint: Draw the graph of the curve \(y=\frac{1}{x}\) in the \(x y\) -plane. Think of \(\ln n\) as the area under part of this curve. Draw appropriate rectangles above the curve.

Express $$ 0.859859859 \ldots $$ as a fraction; here the digits 859 repeat forever.
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