Find the sum of 5 + 6 + 8 + 9 + 11 + 12 + 14 + 15 + ... + 998 + 999.

(Hint: Break into two separate series and use properties of summation.)

Note that each term is either 1 or 2 more than the previous term (it alternates), so we can split the series up as follows:
5 + 6 + 8 + 9 + 11 + 12 + 14 + 15 + ... + 998 + 999
= (5 + 8 + 11 + 14 + ... + 998) + (6 + 9 + 12 + 15 + ... + 999).

These are both arithmetic series. The first series has a first term of 5, a last term of 998, and has 332 terms. The second series has a first term of 6, a last term of 999, and also has 332 terms. Since the sum of the first n terms of an arithmetic series, S_n, with a first term of a₁ and a last term of a_n is:
S_n = (n/2)(a₁ + a_n),

we see that:
(a) 5 + 8 + 11 + 14 + ... + 998 = (332/2)(5 + 998) = 166498
(b) 6 + 9 + 12 + 15 + ... + 999 = (332/2)(6 + 999) = 166830.

Therefore, the series has sum 166498 + 166830 = 333328.

I hope this helps!

two seperate series:
5+11+8...998
and
6+9+12...999

the number of terms in series 1 is (998-5)/3+1=332 terms
the number of terms in series 1 is (999-6)/3+1=332 terms

(5+998)(332/2)=166498
because 332/2 is the number of pairs and 5+998 has the same sum as 11+992 and so on)

(6+999)(332/2)=166830

add the two series together to get the total sum:
333,328
ta-da that's the answer

series 1: 5, 8, 11, 14, 17...998
series 2: 6, 9, 12, 15, 18...999

series 1: a1 = 5, d = 3, n = 332
series 2: a1 = 6, d = 3, n = 332

sum of a series with n terms: Sn = (n/2)(a1 + an)
sum of series 1: Sn = (332/2)(5 + 998)
= 166 * 1003
= 166,498

sum of series 2: Sn = (332/2)(6 + 999)
= 166 * 1005
= 166,830

Add the two sums together, and you get 333,328.

5 + 6 + 8 + 9 + 11 + 12 + 14 + 15 + ... + 998 + 999
= (5 + 8 + 11 + 14 + ... + 998) + (6 + 9 + 12 + 15 + ... + 999)
= 166 498 + 166 830
= 333 328

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