Evaluate the Summation sum from k=5 to 9 of sin(kpi)
The question is asking for the calculation of a mathematical sum where the summation process involves taking the sine of each integer value of k multiplied by π (pi), starting from k = 5 and ending at k = 9. The summation would require the evaluation of sin(5π), sin(6π), sin(7π), sin(8π), and sin(9π), and then adding up all these values to find the total sum.
$\sum_{k = 5}^{9} sin \left(\right. k \pi \left.\right)$
Answer
Solution:
Evaluate the Summation $\sum_{k=5}^{9} \sin(k\pi)$
Step 1:
Write out the summation explicitly for each integer $k$ from 5 to 9.
$\sin(5\pi) + \sin(6\pi) + \sin(7\pi) + \sin(8\pi) + \sin(9\pi)$
Step 2:
Calculate each term in the sequence, noting that the sine of any integer multiple of $\pi$ is 0.
$0 + 0 + 0 + 0 + 0 = 0$
Knowledge Notes:
The problem involves evaluating a finite summation of sine functions where the argument of the sine function is an integer multiple of $\pi$. The relevant knowledge points for solving this problem include:
Summation Notation: The summation notation $\sum$ is used to denote the sum of a sequence of terms. The expression $\sum_{k=a}^{b} f(k)$ means that you should evaluate the function $f(k)$ for every integer $k$ from $a$ to $b$, and then add up all those values.
Sine Function Properties: The sine function has a period of $2\pi$, which means that $\sin(\theta + 2\pi n) = \sin(\theta)$ for any integer $n$. Moreover, for any integer $m$, $\sin(m\pi) = 0$ because sine is zero at multiples of $\pi$.
Simplification of Series: When evaluating a series, if each term in the series simplifies to zero, the sum of the series is also zero.
In this problem, we use the property of the sine function that $\sin(m\pi) = 0$ for any integer $m$ to simplify each term in the summation to zero. Since all terms are zero, the sum is zero.
