Evaluate the following summation (make sure your solution is in radians): ∑ k = 0 5 2 tan ( π k 3 ) − π k \sum_{k=0}^52\tan\left(\frac{\pi k}{3}\right)-\pi k

Let's give this problem a try. So here we're asked to evaluate the following summation and to make sure our solution is in radians. Now this looks kinda scary at first because we're dealing with this tangent, and then we see these pis in here, but don't sweat it because it turns out we can apply the same rules that we've learned for summations to this problem as well. Now what I'm first going to do is take this and split it into two separate sums. So since I see we have two times the tangent of pi k over three, I'm gonna make that one sum right here. So we're gonna have from p equals zero to five of this whole portion. So we'll have two times the tangent of pi k over three. Then we're going to have minus, and then we're gonna do the next sum. So we're gonna go from k equals zero to five of this portion, pi times k. So we've now taken this and split it into two separate sums. Now what I can also do is use the constant multiple rule since we're dealing with these summations. So any constants that I see can come in front of the summation symbol. So in this case, we'll have two, which is gonna come out the front of this sum. So we'll have from k equals zero to five for the tangent of pi k over three. Then we're going to have minus, and then we're going to have pi, which we can bring out to the front because pi is a constant in this problem. So we're gonna have from k equals zero to five for k. And pi is always this constant that you can keep in mind. So it's not like some variable. We can bring it out to the front. So this is what happens when we do this. Now let's go ahead and evaluate these sums. I'll start with this sum right here, where we have the summation, which goes from k equals zero to five of the tangent right here, tangent of pi k over three. Now what I need to do is start with zero and and plug it in for k and then keep incrementing up. If I plug zero in for k, everything here is just gonna become zero. So we're gonna have the tangent of zero plus, and then we'll have k equals one that we plug in. So we're going to have the tangent and if we plug one in for k, we're just gonna get the tangent of pi over three because that's what happens when you multiply it by one. And then we're going to have plus, and then we'll have the tangent, and then I'll go ahead and plug in two because we're incrementing up till we get to five. So we'll have two pi over three. And then we're going to have plus and we're just gonna keep going so we'll have the tangent and then the next one would be three pi over three which is just which is just gonna be pi so we're gonna have the tangent of pi and we're gonna have plus the tangent of four pi over three So plus the tangent of five pi over three. And then this is going to be what happens when we expand this all out. Now it turns out you can take these numbers right here, and you can plug this all into a calculator. And if you do this, you'll notice an interesting result. This all just comes out to zero. Doing this sum is just gonna give us zero when we get the output as our result. So this is what the entire sum becomes. So that means this whole thing is just going to become two times zero when we do this. Now next, I'm going to deal with this portion of the sum, which is from zero to five four k. So to do this, we'll have from k equals zero to five, and then we're dealing with k here. So what this is going to be is zero plus one plus two plus three plus four plus five. And what happens is we can add all these numbers up, and what we should get is a result of 15. Because we have one plus two, which is three, three plus three is six, plus four that gives us 10, plus five gives us this 15, that plus zero that doesn't do anything. So we just end up with 15, meaning this whole thing is going to become minus pi times 15. So this is what we end up getting. And from here, what we can do is simplify all of this to just get negative 15 pi as our result since that zero is just gonna go away. The whole thing becomes zero. So negative 15 pi, which on a calculator comes out to negative 47.12 radians is going to be the result. So either one of these would be the correct solution to the problem, and that is how you can solve things. So this would be our solution. That's how you can deal with sums when dealing with trigonometric functions or pi when you see them. So hope you found this video helpful, and let's move on.

8. Definite Integrals

Riemann Sums

Calculus

8. Definite Integrals

Riemann Sums

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!

Multiple Choice

Evaluate the following summation (make sure your solution is in radians):
$\sum_{k = 0}^{5} 2 \tan (\frac{π k}{3}) - π k$

-43.65

-47.12

-50.58

Verified step by step guidance

Identify the summation expression: $ \sum_{k=0}^{5} \left( 2 \tan\left(\frac{\pi k}{3}\right) - \pi k \right) $. This means you will evaluate the expression inside the summation for each integer value of $ k $ from 0 to 5 and then sum the results.

For each term in the summation, calculate $ \tan\left(\frac{\pi k}{3}\right) $. Remember that the tangent function is periodic with a period of $ \pi $, and you should use radians for the angle.

Multiply the result of the tangent function by 2 for each $ k $, as indicated by the expression $ 2 \tan\left(\frac{\pi k}{3}\right) $.

Subtract $ \pi k $ from the result obtained in the previous step for each $ k $. This gives you the value of the expression $ 2 \tan\left(\frac{\pi k}{3}\right) - \pi k $ for each $ k $.

Sum all the values obtained from $ k = 0 $ to $ k = 5 $ to get the final result of the summation. This involves adding up all the individual results from the previous step.

4:22m

Watch next

Master Sigma Notation with a bite sized video explanation from Patrick

Riemann Sums practice set

Problem sets built by lead tutors

Expert video explanations

Struggling with Calculus?

Evaluate the following summation (make sure your solution is in radians):∑k=052tan(πk3)−πk\(\sum\)_{k=0}^52\(\tan\[\left\)(\(\frac{\pi k}{3}\]\right\))-\(\pi\) k

Watch next

Riemann Sums practice set

Evaluate the following summation (make sure your solution is in radians):
$\sum_{k = 0}^{5} 2 \tan (\frac{π k}{3}) - π k$