Question 898860
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Let the first odd integer be 2k-1 where k is an integer

The sum of an arithmetic series is

{{{S[n]=expr(n/2)(2a[1]+(n-1)d)}}}

With odd integers d=2, we'll let a<sub>1</sub> = 2k-1, n=384

{{{S[384]=expr(384/2)(2(2k-1)+(384-1)2[""])}}}

{{{S[384]=192(2(2k-1)+(383)2[""])}}}

{{{S[384]=2^6*3(4k-2+766)}}}

{{{S[384]=2^6*3(4k+764)}}}

{{{S[384]=2^6*3*4(k+191)}}}

{{{S[384]=2^6*3*2^2*(k+191)}}}

{{{S[384]=2^8*3(k+191)}}}

{{{2^8}}} is a perfect 4th power. Since we have one factor of 3,

the smallest 4th power possible will be {{{2^8*3^4}}}.

That's the answer: 

{{{2^8*3^4=256*81=20736=12^4}}}, choice b

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Following through,

This will be when {{{k+191 =3^3}}} 

{{{k+191 =27}}

{{{k=-164}}}

Notice that the problem just said "consecutive odd integers", it didn't
say they had to be positive!

So the sequence of 384 odd integers starts with 2k-1 = 2(-164)-1 = -329

and ends with {{{a[384]=-329+(384-1)(2)=437}}}

So it's the series of 384 consecutive odd integers

(-329)+(-327)+...+435+437 = 20736 = 12<sup>4</sup>

Edwin</pre>