Question

find the sum of the finite arithmetic series.\\(\sum_{n = 0}^{18} (4n + 4)\\)

Explanation:

Step1: Recall the formula for the sum of an arithmetic series

The sum of an arithmetic series \(\sum_{n = 0}^{k} a_n\) where \(a_n=a_1+(n - 1)d\) (or in our case, we can also use the formula \(S=\frac{n(a_1 + a_n)}{2}\), where \(n\) is the number of terms, \(a_1\) is the first term, and \(a_n\) is the last term. First, we need to find the number of terms, the first term, and the last term.

The number of terms \(N\) when \(n\) ranges from \(0\) to \(18\) is \(18 - 0+ 1=19\) terms.

Step2: Find the first term (\(a_1\)) when \(n = 0\)

Substitute \(n = 0\) into \(a_n=4n + 4\):
\(a_0=4(0)+4 = 4\)

Step3: Find the last term (\(a_{18}\)) when \(n = 18\)

Substitute \(n = 18\) into \(a_n=4n + 4\):
\(a_{18}=4(18)+4=72 + 4=76\)

Step4: Use the sum formula for arithmetic series \(S=\frac{N(a_1 + a_N)}{2}\)

Here, \(N = 19\), \(a_1=a_0 = 4\), \(a_N=a_{18}=76\)
\(S=\frac{19\times(4 + 76)}{2}\)
First, calculate the sum inside the parentheses: \(4+76 = 80\)
Then, \(S=\frac{19\times80}{2}\)
\(19\times80 = 1520\), and \(\frac{1520}{2}=760\)

Answer:

\(760\)