求和:Sn=1*2*3+2*3*4+……+n(n+1)(n+2)

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/06 01:18:41
求和:Sn=1*2*3+2*3*4+……+n(n+1)(n+2)
xN0_l⶗H84ē1!C ( D c81Η1WG[o00Y# w/HP!pz^[!S;J=mށ `[[k`0Էb[fyM-e1\Ы%Uۣ uRѲB!BZ 0 B)/o T4p&TZcTyβx||_j\u] K"ǚ&(ph*9 A ʬLA$hxE^#G\UŅ //>^

求和:Sn=1*2*3+2*3*4+……+n(n+1)(n+2)
求和:Sn=1*2*3+2*3*4+……+n(n+1)(n+2)

求和:Sn=1*2*3+2*3*4+……+n(n+1)(n+2)
Sn=1*2*3+2*3*4+……+n(n+1)(n+2)
=(1³+2³+3³+----+n³)+3(1²+2²+3²+---+n²)+2(1+2+3+----+n)
=n²(n+1)²/4+n(n+1)(2n+1)/2+n(n+1)
=n(n+1)[n(n+1)/4+(2n+1)/2+1)]
=n(n+1)(n+2)(n+3)/4

an=n(n+1)(n+2)
=((n+3)-(n-1))/4*n(n+1)(n+2)
=[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]/4
sn所有的项都如此裂项分解,加起来,可以狂消
Sn=1*2*3+2*3*4+3*4*5+...+n*(n+1)*(n+2)
=1/4{1*2*3*(4-0)+2*3*4*(5-1...

全部展开

an=n(n+1)(n+2)
=((n+3)-(n-1))/4*n(n+1)(n+2)
=[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]/4
sn所有的项都如此裂项分解,加起来,可以狂消
Sn=1*2*3+2*3*4+3*4*5+...+n*(n+1)*(n+2)
=1/4{1*2*3*(4-0)+2*3*4*(5-1)+3*4*5*(6-2)...+n*(n+1)*(n+2)[n+3-(n-1)]}
=[n(n+1)(n+2)(n+3)-0]/4
=n(n+1)(n+2)(n+3)/4

收起