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acmuser

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【答案】应助回帖

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感谢参与,应助指数 +1
zzycyy: 金币+5, ★★★★★最佳答案, 很有帮助,谢谢 2012-05-10 10:56:14
37.5^n + 26.5^n = ((64+11)^n + (64-11)^n)/2^n,
since mod((2k+1)^2,8)=1, n has to be odd number,
use binomial expansion, it is easy to show that
(64+11)^n+(64-11)^n = 128m + 128*n*11^(n-1),
therefore, n<=7, check all odd number <=7, we have
n=0,1,3,5,7.
11楼2012-05-10 00:15:25
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到时候

银虫 (小有名气)

原式写成 (32+5.5)^n+(32-5.5)^n,若n为奇数,展开式中5.5的奇数项会抵消,考虑5.5的幂次为n-1的那一项,此时还有一个32和n你为奇数,32中有五个2,要使原式为整数,则n-1<=5+1(加1是因为奇幂次项抵消,偶幂次项加倍,式中2的个数加1),得n<=7;若n为偶数,考虑最后一项,为2*5.5^n,则n只能取0,综上,n可取 0,1,3,5,7.
小楼一夜听春雨
12楼2012-05-10 23:27:21
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xiaoxiao7661

银虫 (小有名气)

n=0,1……n,初中时学过归纳法,这个可以算做一个证明题
13楼2012-05-11 16:48:30
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ccyzrb

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0,1,3,5,7,9,11
14楼2012-07-12 13:10:01
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