Term Calculation for k=0: The series is a finite sum with terms defined by the expression (2k-1) for each integer value of k from 0 to 1. We will calculate each term separately and then add them together. Term Calculation for k=1: First, we calculate the term for k=0. Substituting k=0 into the expression (2k-1) gives us (2*0-1), which simplifies to -1. Sum of Terms for k=0 and k=1: Next, we calculate the term for k=1. Substituting k=1 into the expression (2k-1) gives us (2*1-1), which simplifies to 1. Simplified Sum: Now, we add the two terms together. The sum of the terms for k=0 and k=1 is (-1) + (1). The sum (-1) + (1) simplifies to 0.

Calculus

Find derivatives using the chain rule I

Full solution

\sum_{k=0}^{1}(2 k-1)=

Term Calculation for $k=0$ : The series is a finite sum with terms defined by the expression $(2k-1)$ for each integer value of $k$ from $0$ to $1$ . We will calculate each term separately and then add them together.
Term Calculation for $k=1$ : First, we calculate the term for $k=0$ . Substituting $k=0$ into the expression $(2k-1)$ gives us $(2\times 0-1)$ , which simplifies to $-1$ .
Sum of Terms for $k=0$ and $k=1$ : Next, we calculate the term for $k=1$ . Substituting $k=1$ into the expression $(2k-1)$ gives us $(2\times 1-1)$ , which simplifies to $1$ .
Simplified Sum: Now, we add the two terms together. The sum of the terms for $k=0$ and $k=1$ is $(-1) + (1)$ .
Simplified Sum: Now, we add the two terms together. The sum of the terms for $k=0$ and $k=1$ is $(-1) + (1)$ . The sum $(-1) + (1)$ simplifies to $0$ .