Answered

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A series of 384 consecutive odd integers has a sum that is a perfect fourth power of a positive
integer. Find the smallest possible sum for this series.
A) 104 976 B) 20 736
C) 10 000 D) 1296 E) 38 416

Sagot :

Using an arithmetic sequence, it is found that the smallest possible sum for the series is of 20 736, given by option B.

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  • In an arithmetic sequence, the difference of consecutive terms is always the same, called common difference.
  • The nth term of an arithmetic sequence is given by:

[tex]a_n = a_1 + (n-1)d[/tex]

  • The sum of the first n terms is given by:

[tex]S_n = \frac{n(a_1+a_n)}{2}[/tex]

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  • The set of odd integers is an arithmetic sequence with common difference 2, thus [tex]d = 2[/tex].
  • 384 terms, thus [tex]n = 384[/tex]
  • The last term is: [tex]a_{384} = a_1 + 383(2) = a_1 + 766[/tex]

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The sum of the 384 terms is:

[tex]S_{384} = \frac{384(a_1 + a_1 + 766)}{2} = 192(2a_1 + 766) = 384a_1 + 147072[/tex]

Now, for each option, we have to test if it generates an odd [tex]a_1[/tex].

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Option d: Sum of 1296, thus, [tex]S_{384} = 1296[/tex], solve for [tex]a_1[/tex]

[tex]384a_1 + 147072  = 1296[/tex]

[tex]a_1 = \frac{1296 - 147072}{384}[/tex]

[tex]a_1 = -379.6[/tex]

Not an integer, so not the answer.

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Option c: Test for 10000.

[tex]384a_1 + 147072  = 10000[/tex]

[tex]a_1 = \frac{10000 - 147072}{384}[/tex]

[tex]a_1 = -356.9[/tex]

Not an integer, so not the answer.

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Option b: Test for 20736.

[tex]384a_1 + 147072  = 20736[/tex]

[tex]a_1 = \frac{20736 - 147072}{384}[/tex]

[tex]a_1 = -329[/tex]

Integer an odd, thus, option b is the answer.

A similar problem is given at https://brainly.com/question/16720434

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