{Use of Tech} Sum of squared integers Let T (n) = 1² + 2²...

Question

{Use of Tech} Sum of squared integers Let T (n) = 1² + 2² + ... + n², where n is a positive integer. It can be shown that T (n) = n (n + 1) (2n + 1) / 8

c. What is the least value of n for which T(n) > 1000?

Accepted Answer

Welcome back, everyone. Determine the least value of N for which the sum of cubes of the first N natural numbers is less than 2500 using the following formula. 1 cubed plus 2 cubed plus 3 cubed and so on plus n cubed equals and multiplied by M + 1 divided by 2 squared and we're given 4 answer choices A says 8, B 9, C 10, and D 11. So we're going to take the general formula. It says N multiplied by N + 1. We are dividing that by 2 and squaring. According to the context of the problem, we want to get a result that is less than 2500. So first of all, what we're going to do is basically take square root of both sides, and we're only going to get one solution for the left hand side because N is positive, right? So when we do that, we get N multiplied by N + 1 divided by 2 is less than. By less than 50, right? Now we can multiply both sides by 2, and essentially we can remove 2 from the denominator and on the right hand side we're going to get 100. So now let's rearrange this into a quadratic inequality. We're going to get n2 plus N minus 100. We're subtracting 100 from both sides. This must be less than 0. So first of all, let's identify the zeros of this. Parabola, right? Let's assume that this is a parabola. It has a leading coefficient of 1, which is positive. So essentially we can say that this parabola opens up. If we manage to solve an inequality where it is less than 0, well, we are essentially looking at the region below the x axis. So let's solve the equation in the form of n2 plus N minus 100 is equal to 0, and we can use the quadratic formula. So we will have 2 solutions and 1 and and 2. Essentially, we have to use the formula which says that. We take minus B plus or minus square root of the discriminant B2 minus 4AC, and then we are dividing by 2A. And in our case, A is the leading coefficient. It's 1. B is the coefficient in front of N. It is also 1, and C is the free coefficient, which is 100. So we get -1 plus or minus square root of. Now B2 gives us a 12 minus 4AC. So 4 multiplied by 1 multiplied by -100, and we are dividing that by 2A, so 2 multiplied by 1. And we get two solutions. Now for the first one, we're going to get -1. Plus square root of 401 divided by 2. Now, let's approximate this result. We're going to get approximately 9.5. For the second solution, it is going to be negative. Well, we cannot take a negative solution, right, because N must be positive. We start with N equals 1. It is a natural number. So we're only interested in the first solution, which is 9.5 approximately, right? And essentially, if we are looking for a natural number. To satisfy the region below the x axis since this parable opens up, well, we are interested in the nearest natural number, but essentially we have to round down right to make sure that we're still in this region. So we can say that N is equal to 9, right? And this means that the correct answer to this problem corresponds to the answer choice B. Thank you for watching.

{Use of Tech} Sum of squared integers Let T (n) = 1² + 2² + ... + n², where n is a positive integer. It can be shown that T (n) = n (n + 1) (2n + 1) / 8c. What is the least value of n for which T(n) > 1000?

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{Use of Tech} Sum of squared integers Let T (n) = 1² + 2² + ... + n², where n is a positive integer. It can be shown that T (n) = n (n + 1) (2n + 1) / 8

c. What is the least value of n for which T(n) > 1000?