Question

n^n+1/(n+1)^n+2怎么算

Answer 1

1/n(n+1)(n+2)求和怎么做1/n(n+1)(n+2)
=1/2[2/n(n+1)(n+2)]
=1/2[(n+2)-n]/n(n+1)(n+2)]
=(1/2)[(n+2)/n(n+1)(n+2)-n/n(n+1)(n+2)]
=(1/2)[1/n(n+1)-1/(n+1)(n+2)]
所以和=(1/2)[1/1*2-1/2*3+1/2*3-1/3*4+……+1/n(n+1)-1/(n+1)(n+2)]
=(1/2)[1/1*2-1/(n+1)(n+2)]
=n(n+3)/[4(n+1)(n+2)]
1/n（n-1）（n+2）怎么求和分母是n(n+1)(n+2)
1/n(n+1)(n+2)=1/2·2/n(n+1)(n+2)
=1/2·[(n+2)-n]/n(n+1)(n+2)
=1/2·[(n+2)/n(n+1)(n+2)-n/n(n+1)(n+2)]
=1/2·[1/n(n+1)-1/(n+1)(n+1)]