91Ó°ÊÓ

An arithmetic series is a sum of terms from an arithmetic sequence which has a constant difference between each term. To calculate the sum of the first \(n\) terms of an arithmetic series, we use a specific formula. For a series where the first term is \(a\) and the last term is \(T_n\), the sum \(S_n\) is given by:
\[S_n = \frac{n}{2}(a + T_n)\]The formula essentially averages the first and last terms and multiplies by the number of terms. In our exercise, the series is defined by the general term \((5n - 4)\). Using this, after determining that the first term is 1 and difference between terms is 5, we substitute the \(n\)-th term to find the sum:

• \(a = 1\)
• \(T_n = 5n - 4\)

We plug these into the sum formula:
\[S_n = \frac{n}{2}(1 + 5n - 4) = \frac{n(5n - 3)}{2}\]This calculation convinces us that the sum formula is indeed what the exercise claims, valid for all integers \(n \geq 1\).

The general term of an arithmetic sequence provides a quick way to calculate any term in the sequence without listing them all. For an arithmetic sequence, the general term \(T_n\) can be expressed using the formula:
\[T_n = a + (n-1)d\]where \(a\) is the first term and \(d\) is the common difference. In our example, the first term \(a\) is 1, and the common difference \(d\) is 5.Therefore, the formula for the general term of this series becomes:
\[T_n = 1 + (n-1) \, 5 = 5n - 4\]This formula generates each term of the series directly. By plugging in different values of \(n\), you can determine any term in the sequence, making computations and checks straightforward and efficient.

Proof by induction is a powerful method in mathematics for demonstrating the truth of an infinite sequence of statements. It involves two basic steps:

• Base Case: Verify that the statement is true for the initial value, often \(n=1\).
• Inductive Step: Assume the statement holds for \(n=k\), and then prove it also holds for \(n=k+1\).

In the context of our series, the base case requires checking that the formula \(\frac{n(5n-3)}{2}\) correctly computes the sum for the initial term, which it does as you can verify by computing directly.For the inductive step, assume the formula holds for \(n=k\):
\[S_k = \frac{k(5k-3)}{2}\]Then prove for \(n=k+1\), expressing \(S_{k+1}\) as:
\[S_{k+1} = S_k + T_{k+1}\]Using the general term \(T_{k+1} = 5(k+1)-4\), substitute and simplify, thus confirming that the formula of \(S_{k+1} = \frac{(k+1)(5(k+1)-3)}{2}\) holds.This thorough method ensures that for every integer \(n\geq1\), the sum formula accurately reflects the series sum.