自然数N次方求和公式推导

试推导

aaaai12ni1n公式

第一步：

na(n1)an[n(1)CnCn

1aa1aa1aa1(1)Cnmmmaam(1)a1Ca1an(1)]a(1)Cn2aa2(1)Cm1am1an(1)a2Ca1an(1)第二步：

1a2ana[n(n1)]1a[(n1)(n2)]2a[(nm1)(nm)]ma(21)(n1)ana1an(2a1a)(n1)(3a2a)(n2)[(m1)ama](nm)[na(n1)a][nn[1a2a1a3a2ana(n1)a]Cna1C

1an1a(n1)2a(n1)2a(n2)ma(nm1)ma(nm)2(n1)a(n1an(2a1a)n1(2a1a)(3a2a)n2(3a2a)[na(n1)a]n(n1)[na(n上式中C1(2a1a)2(3a2a)m[(m1)ama](n1)[na

由第一步可得

1a12a1aCa2(1)Ca22a2(1)mCam12am1(1)a2Caa12(1)a132C3aa1a1a(1)C32a2a(1)Cmm1am1a3(1)a2C3(1)a1aa11a1na(n1)aCan(1)Ca2na2(1)mCam1nam1(1)a2Caa1n(1)a1

第三步：

1a12a2m1am1a1C1[Ca2(1)Ca2(1)mCa2(1)a2Ca2(1)a1]2[C31a1a(1)C32aa2(1)Cmm1am1a3(1)a2Ca1a3(1)a1]1a12a2m1am1a1(n1)[Can(1)Can(1)mCan(1)a2Can(1)a1]12Ca[12a123a1(n1)na1](1)Ca[12a223a2(n1)nam1a1(1)mCa[12am123am1(n1)nam1](1)a2Ca[122a(1)a1Ca[12(n1)]

1Ca[(1a1a1)(2a2a1)(nana1)](1)Ca2[(1a11a2)(2a12a2)(na1(1)mCam1[(1am1am1)(2am2am1)(namnam1)](1)a2Caa1[(121)(222)(n2n)](1)a1[12(n1)]1Ca[iaia1](1)Ca2[ia1ia2](1)mCam1[iamiam1]i1i1i1i1i1i1a2nnnnnn(1)nC[ii](1)a1a2i1i1nnna1i(1)nai1nni1i1n3a2mm1amaia(1)Ca21ia1(1)2Cai(1)Ci1a1i1i112a1aa(1)a2Caai(1)Ci(1)n1a1i1i1nn

第四步：将第三步C值代入第二步得

aa1a2a123a2in[ai(1)Ci(1)Cia1a1i1i1i1i1nnnnnnnm1ama2a12a1aa(1)mCai(1)Ci(1)Ci(1)n]1a1a1i1i1i1

2a13a2(a1)iana1Cai(1)Ci1a1i1i1i1mm2a1nnn(1)C

ii1nam1(1)a3Ca1a1i(1)2i1na2Caa1a1i(1)ni1n1a1Cnaaia12i1nmm1an2Ca1ai(1)3i1nii1na2na1aCam1a3C2a2C(1)i(1)i(1)m2i1a1i1a

a2aii1n(

利用此公式可得

32n(n1)(2n1)2n3nn2i66i1n2432[n(n1)]n2nn3i44i1n2543n(n1)(2n1)(3n3n1)6n5n10nn4i3030i1 n2226542n(n1)(2n2n1)2n6n5nn5i1212i1n437653n(n1)(2n1)(3n6n3n1)6n21n21n7nn6i4242i1n2243287642n(n1)(3n6nn4n2)3n12n14n7n2n7i2424i1n

利用公式可得：

na2C0(1)i(1)a12i1a1n1aa13Ci(1)3i1n2aa2ii1nm1na2na1na1CCC(1)nmam1a1a2aaa(1)i(1)i(1)imi1a1i1ai1a10na1(1)Cn1a1ii1na(1)C22a1ii1na1(1)C33a1ii1ni1na212aaa1(1)mCam1iam1(1)a1Caai(1)Ci(1)n1a1i1i1n

令a+1=p，则有

nCp1pii1np1(1)Cn2pii1np2(1)Cn23pii1np3(1)Cmm1ppm1ii1np22p2p1p1(1)p3Cpi(1)Ci(1)npi1i1 …………（1）

另外

n[(n1)1]pp

P12P2mPmp22p1np(n1)pC1(n1)C(n1)C(n1)C(n1)Cppppp(n1)1P1P12P2P2np(n2)pC1[(n1)(n2)]C[(n1)(n2)]ppmp1Cp[(n1)Pm(n2)Pm]Cp[(n1)(n2)]2P1P1P1P12P2P2P2P2np1pC1[(n1)(n2)21]C[(n1)(n2)21]ppmp1Cp[(n1)Pm(n2)Pm2Pm1Pm]Cp[(n1)(n2)21](n1)

nCp1pii1n1p1C2pii1n1p2Cmpii1n1pmCp1pini1n1 ………………………

…………（2）

p12p2mpmp11p12p2mpmnpC1nCnCnCnCiCiCipppppppi1i1i1nnn

aaaa(123n)的另一递推式：

由（2）式可得出计算

1a1C(n1)aai[a12i1na2a1nCaC2aia1i1ana1ii1n2Ca3a2ii1nmnCaamim1i1n1i]a1i1

（1）式加（2）式得到

p1p12p2p232p3p2npC1(ii)C[(1)ii]C[(1)iipppi1i1i1i1i1i1m1p2Cp[(1)mipm1ipm1]Cp[(1)p3i2i2]i1ni1i1i1p1Cp[(1)p2ii][(1)p1nn]i1i1n1nn1nn1nn1nn1nn1

（1）当p为奇数时，有

p4nnnnnp1p3(i)()(i)()p2223i1242i1nnpp1C1p1p22Cp1p3C3p1

2k2Cp12k1(ii1np2k1np2k12)2k1Cp12k(np2kn)(i)2p2i12p2p3Cp1n2pCp1（2）当p为偶数时，有

pnnnnnp1p3(i)()(i)(p2223i1242i1nnpp1C1p1p22Cp1p3C3p12

C2k2p12k1(ii1np2k1np2k12)C2k1p12k(np2kCn)()(2p22p1Cp3p1p2p1

令p-1=a，由上面两式可得到下面两式：

（1）当a为偶数时，有

123a1a2a3nCCCnnnaa22anaani(1)(i)(1)a12223i1242i1na1a(1)(1)

2k1Cnna2k2kC(i)(1)2k1i122k22a1Canna1(1)a2a12kana2k2k1aa2k1(1)a3C(a1ia2aa2（2）当a为奇数时，有

a3CCCnnnnn2a2a)(1)(i(1)i4223i122a12i1na1aa1na21a2a3a(1)2k1a2k12ka2k1a2knCCCanna32ka2kaa(1))(1)(ia2k2222k1i1(1)a2Ca1naa(ini12)

从上面两式可以看出：（1p+2p+…+np）展开式中无na-2、na-4、na-6……项。事实上，nia2i1中最

na1na2高项为a1，第二项为2，减去括号中后一项后即无na-2项。其余同理。