No. There is no number n, such that the sum of the first n consecutive integers is 1,000. Here's why: Let S = S(n) = 1 + 2 + 3 + . . . + n. Then, S = n(n + 1) / 2. There are several ways you can derive this formula; but you might find it helpful to refer to this: http://mathworld.wolfram.com/ArithmeticSeries.html or any of a large number of similar sources. (I tried Wiki, but down-loading took forever!) If you know S(n) (in this case, 1,000) and wish to find n, you have only to solve: n² + n = 2S = 2000. This is not readily factorised; but we can try completing the square: 2S = n² + n = 2,000; whence, n² + n + ¼ = (n + ½)² = 2,000¼, and n = √2,000.25 - 0.5. To the nearest integer, n = 44, approximately; but this is inexact, of course. Let's see what happens when we try this result. We get 2S(44) = n² + n = 44 * 45 = 1,980; whence, S(45) = 990, which is just a little too small; and any n smaller than 44 would give an S(n) that is even smaller. You can easily check this result, if you like: 1 + 2 + 3 + . . . + 22 + 23 + . . . + 42 + 43 + 44 = (1 + 44) + (2 + 43) + (3 + 42) + . . . + (22 + 23) {this makes 22 pairs, each adding up to 45.} = 45 * 22 = 990. On the other hand, S(45) = S(44) + 45 = 990 + 45 = 1,035, which is just a little too large; and any n larger than 45 would give an S(n) that is even larger. We may conclude, then, that no integer n satisfies the requirement that we have placed on it.
Basically, a knowledge of even and odd functions can simplify certain calculations. One place where they frequently appear is when using trigonometric functions - for example, the sine function is odd, while the cosine function is even.
The first 10 digits are 3.1415926535 and these are sufficient for all but the most rigorous calculations. There is a text file available at the related link that has the first billion (takes at least 35 seconds to load).The field size of this page cannot accommodate even the first 200,000 digits.
An even function is symmetric around the vertical axis. An odd function - such as the sine function - has a sort of symmetry too - around the point of origin. If you graph this specific function (for example, on the Wolfram Alpha website), you can see that the function has none of these symmetries. To prove that the function is NOT even, nor odd, just find a number for which f(x) is neither f(-x) nor -f(-x). Actually proving that a function IS even or odd (assuming it actually is) is more complicated, of course - you have to prove that it has the "even" or the "odd" property for EVERY value of x. Let f(x) = 2x3 - x2. Notice that f is defined for any x, since it is a polynomial function. If f(-x) = f(x), then f is even. If f(-x) = -f(x), then f is odd. f(-x) = 2(-x)3 - (-x)2 = -2x3 - x2 Since f(-x) ≠ f(x) = 2x3 - x2, f is not even. Since f(-x) ≠ - f(x) = -(2x3 - x2) = -2x3 + x2, f is not odd. Therefore f is neither even nor odd.
Geometry is used in many different ways in real life. For example, if you wanted to measure the volume of a circle so that you could know beforehand if some liquid you wanted to get into it would all fit, you could find out beforehand; geometry is used for measurements of things as small as atoms or cells to the size of the earth (and maybe even further)...eventually, you will find that it was great to learn geometry.
The sum of the squares of two consecutive positive even integers is 340. Find the integers.
There are no two consecutive even integers, consecutive odd integers, or consecutive integers that satisfy that relationship.
Find three consecutive positive even integers whose sum is 123 , Answer
No. There is no set of three consecutive even integers with a sum of 40.
The numbers are 82, 84, 86 and 88.
The even integers are 22, 24 and 26.
The two consecutive, even integers are 350 and 352.
The integers are -37, -36 and -35. Also, using consecutive even integers: -38, -36 and -34.
The sum of two consecutive odd integers is an even number. There is no possible answer to this question.
The three integers are 25, 26, and 27.
No.The sum of any number of even integers, consecutive or not, MUST be even. 219 is not even.No.The sum of any number of even integers, consecutive or not, MUST be even. 219 is not even.No.The sum of any number of even integers, consecutive or not, MUST be even. 219 is not even.No.The sum of any number of even integers, consecutive or not, MUST be even. 219 is not even.
12 and 14.