 In this video we provide the solution to question number eight for the practice final exam for Maths 1210. We're asked to find the sum where k ranges from one to fourteen of the terms k squared minus four. So if I was approaching this one, I would break it up into two pieces. Let's take the sum of the k squares minus the sum of the fours, again as we're ranging from one to fourteen here, equals one to fourteen. For which when you take the sum of the squares we're going to apply the formula of that situation, which is n times n plus one times two n plus one all over six. And we're doing this in the instance where n equals, sorry in this case fourteen. And then for the other one that's a lot easier, you're going to get four times n. But in that case n again is fourteen. So again much easier to do. So plugging these values into the formulas that we have here we're going to have fourteen times fifteen times twenty nine over six. And then we're going to subtract from the next one four times fourteen. For which I'm going to actually simplify the fraction before I multiply things out because sixteen of course is just two times three. Two goes into fourteen seven times three goes into fifteen five times. And so with that in mind we're going to take seven times five times twenty nine which is going to give us one thousand and fifteen. We also have four times fifteen which is fifty six and then excuse me fifty six. And of course if we take one thousand fifteen and subtract from a fifty six we're going to end up with nine hundred and fifty nine. In which case we see the correct answer is C.