1999年5月7日 · 1000000 This series has (1999-1)/2+1 = 1000 terms. The average term is the same as the average of the first and last elements, namely (1+1999)/2 = 1000 So the sum is 1000*1000 = 1000000

2016年9月11日 · T_n = 2n -1 This is clearly an arithmetic sequence because the terms differ by 2 each time. To find the nth term rule we need: a value for the first term , a and a value for the common difference d .

2017年9月20日 · Assuming that the question is correct as given, there are several way to match the sequence: 1,3,7,9 with a formula. For example, if we add (− 1)n 2 to each element, then we get the sequence: 1 2, 7 2, 13 2, 19 2 which is an arithmetic sequence with common difference 3. Hence we can write a formula for the given sequence: an = 3n − 5 2 − (− 1)n 2 Alternatively, we could use the method ...

An arithmetic sequence is a sequence (list of numbers) that has a common difference (a positive or negative constant) between the consecutive terms. Here are some examples of arithmetic sequences: 1.)7, 14, 21, 28 because Common difference is 7. 2.) 48, 45, 42, 39 because it has a common difference of - 3.

2018年4月3日 · Step 4: By proof of mathematical induction, this statement is true for all integers greater than or equal to 1 (here, it actually depends on what your school tells you because different schools have different ways of setting out the final step but you get the gist of it)

2018年2月27日 · Alright, Lets look at this sequence. Is there anything you notice between the first two numbers? How about... 7-5=2 Lets see if this continues to be true 5-3=2 3-1=2 So the pattern is that it is just adding two (or vise versa) to ever number in the sequence. So if we continue it would look like... 11, 9, 7, 5, 3, 1, -1, -3, -5, -7, -9 Notice …

2015年12月7日 · For an arithmetic sequence with initial value a and a difference between terms of d the sum of the first n terms is given by the formula XXXΣ = n 2 ⋅ (2a + (n − 1)d) For the given sequence XXXa = 1 and XXXd = 2 We are told that the required sum is 961 So XXX961 = n 2 ⋅ (2(1) +(n −1)(2)) XXX961 = n 2 ⋅ 2 + n 2 (n −1)(2) XXX961 = n ...

2017年9月6日 · Observ that 1 3 = 1 2(1 − 1 3) 2 15 = 1 2(1 3 − 1 15) 3 105 = 1 2 (1 15 − 1 105) .... In the infinite sum the second term of each item can be simplified with the first term of the following item and the second term goes to 0, so the infinite sum …

2015年9月7日 · This is the sequence of all the odd numbers between 1 and 99, endpoints included. Clearly this is an arithmetic sequence with common difference d = 2 between terms.

2017年8月2日 · Not possible. All the available numbers are odd. The sum of 3 odd numbers must be odd as well. 30 is an even number.