Question

An arithmetic series begins 29+25+21+... find the maximum value that the series could have

Answer 1

The series' value is maximized if the sum only consists of positive terms. Notice that each term in the sum takes the form 4n + 1 for integer n. This means the smallest positive integer that the sum can involve is 1, so the maximum value is

S = 29 + 25 + 21 + … + 9 + 5 + 1

Reversing the order of terms gives the same sum,

S = 1 + 5 + 9 + … + 21 + 25 + 29

Adding terms in the same positions gives us twice this sum,

2S = (29 + 1) + (25 + 5) + (21 + 9) + … + (1 + 29)

Notice how each grouped sum adds to 30. There are 8 terms in the sum, since 4n + 1 = 29 when n = 8. So

2S = 8 × 30 = 240   ===>   S = 120