中等数学2015年增刊模拟题(20)——0

已知正奇数 n n n 满足

n ∣ ∑ k = 1 n k 2014 n\mid \sum_{k=1}^nk^{2014} nk=1nk2014

证明:

n 2 ∣ ∑ k = 1 n k 2015 n^2\mid\sum_{k=1}^nk^{2015} n2k=1nk2015

证:
一般化,将2015改为不小于3的奇数 m m m.
下面证明:若

n ∣ ∑ k = 1 n k m − 1 n\mid\sum_{k=1}^nk^{m-1} nk=1nkm1

n 2 ∣ ∑ k = 1 n k m n^2\mid\sum_{k=1}^nk^m n2k=1nkm

为此,设
∑ k = 1 n k m − 1 = n q . \sum_{k=1}^nk^{m-1}=nq. k=1nkm1=nq.

∑ k = 1 n k m = ∑ k = 0 n k m = ∑ k = 0 n ( n − k ) m \sum_{k=1}^nk^m=\sum_{k=0}^nk^m=\sum_{k=0}^n(n-k)^m k=1nkm=k=0nkm=k=0n(nk)m
= ∑ k = 0 n ( n m − C m 1 n m − 1 k + C m 2 n m − 2 k 2 − ⋯ − C m m − 2 n 2 k m − 2 + C m m − 1 n k m − 1 − k m ) =\sum_{k=0}^n(n^m-C_m^1n^{m-1}k+C_m^2n^{m-2}k^2-\cdots-C_m^{m-2}n^2k^{m-2}+C_m^{m-1}nk^{m-1}-k^m) =k=0n(nmCm1nm1k+Cm2nm2k2Cmm2n2km2+Cmm1nkm1km)
= n 2 S + m n ∑ k = 0 n k m − 1 − ∑ k = 0 n k m =n^2S+mn\sum_{k=0}^nk^{m-1}-\sum_{k=0}^nk^m =n2S+mnk=0nkm1k=0nkm
= n 2 S + n 2 m q − ∑ k = 0 n k m =n^2S+n^2mq-\sum_{k=0}^nk^m =n2S+n2mqk=0nkm

2 ∑ k = 1 n k m = n 2 ( S + n q ) . 2\sum_{k=1}^nk^m=n^2(S+nq). 2k=1nkm=n2(S+nq).
于是, n 2 ∣ 2 ∑ k = 1 n k m . n^2\mid2\sum_{k=1}^nk^m. n22k=1nkm.
n n n 为奇数,因此 n 2 ∣ ∑ k = 1 n k m . n^2\mid\sum_{k=1}^nk^m. n2k=1nkm.

你可能感兴趣的:(数学科普)