URGENT! Use Gauss's approach to find the following sums 1+3+5+7+...997Formula please and solve

Answer:The sum of the first 499 terms is 249001Step-by-step explanation:* Lets revise the arithmetic sequence- There is a constant difference between each two consecutive   numbers- Ex:# 2  ,  5  ,  8  ,  11  ,  ……………………….# 5  ,  10  ,  15  ,  20  ,  …………………………# 12  ,  10  ,  8  ,  6  ,  ……………………………* General term (nth term) of an Arithmetic sequence:- U1 = a  ,  U2  = a + d  ,  U3  = a + 2d  ,  U4 = a + 3d  ,  U5 = a + 4d- Un = a + (n – 1)d, where a is the first term , d is the difference  between each two consecutive terms n is the position of the  number- The sum of first n terms of an Arithmetic sequence is calculate from  Sn = n/2[a + l], where a is the first term and l is the last term* Now lets solve the problem∵ The terms of the sequence are 1 , 3 , 5 , 7 , ......... , 997∵ The first term is 1 and the second term is 3∴ The common difference d = 3 - 1 = 2∵ The first term is 1∵ The last term is 997∵ The common difference is 2- Lets find how many terms in the sequence∵ an = a + (n - 1) d∴ 997 = 1 + (n - 1) 2 ⇒ subtract 1 from both sides∴ 996 = (n - 1) 2 ⇒ divide both sides by 2∴ 498 = n - 1 ⇒ add 1 for both sides∴ n = 499∴ The sequence has 499 terms- Lets find the sum of the first 499 terms  ∵ Sn = n/2[a + l]∵ n = 499 , a = 1 , l = 997∴ S499 = 499/2[1 + 997] = 499/2 × 998 = 249001 * The sum of the first 499 terms is 249001