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

Formula please and solve

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Answer:

The sum of the first 499 terms is 249001

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

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