Question

find the sum of the finite series.
\\(\displaystyle\sum_{n=0}^{16} 19(-1)^n\\)

Explanation:

Step1: Analyze the series terms

The series is \(\sum_{n = 0}^{16}19(-1)^{n}\). Let's write out the first few terms. When \(n = 0\), the term is \(19(-1)^{0}=19\times1 = 19\). When \(n = 1\), the term is \(19(-1)^{1}=19\times(-1)= - 19\). When \(n = 2\), the term is \(19(-1)^{2}=19\times1 = 19\), and so on. So this is an alternating series where the terms alternate between \(19\) and \(-19\).

Step2: Determine the number of terms

The number of terms \(N\) in the series \(\sum_{n = 0}^{k}a_{n}\) is \(k + 1\). Here \(k=16\), so the number of terms \(N=16 + 1=17\) terms.

Step3: Group the terms

We can group the terms in pairs (except for the last term if the number of terms is odd). Let's pair the terms as \((n = 0,n = 1),(n = 2,n = 3),\cdots,(n = 14,n = 15)\) and then the last term \(n = 16\).

For each pair \((n = 2m,n = 2m + 1)\) where \(m=0,1,\cdots,7\):
The term at \(n = 2m\) is \(19(-1)^{2m}=19\times1 = 19\)
The term at \(n = 2m + 1\) is \(19(-1)^{2m+1}=19\times(-1)=-19\)
The sum of each pair is \(19+( - 19)=0\)

We have \(7\) such pairs (since \(2\times7 = 14\), so \(m\) from \(0\) to \(7\) gives \(16\) terms), and then we have the 17th term when \(n = 16\).

Step4: Calculate the sum of the pairs and the last term

The sum of the \(7\) pairs is \(7\times0 = 0\). The term at \(n = 16\) is \(19(-1)^{16}=19\times1 = 19\) (since \((-1)^{16}=1\) as \(16\) is even).

So the sum of the series \(S=0 + 19=19\)

Answer:

\(19\)