Question

已知数列{an}的通项公式是an=1/{n(n+2) }(n∈N）,求它的前n项的和。

Answer 1

an=1/n(n+2)={1/n-1/(n+2)}/2

a1+a2+...+an=(1/2)*{1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+...+1/n-1/(n+2)}={1+1/2-1/n-1/(n+2)}/2

Answer 2

因为an=1/2{1/n - 1/(n+2)}
则a1=1/2(1/1 - 1/3)
a2=1/2(1/2 - 1/4)
a3=1/2(1/3 - 1/5)
a4=1/2(1/4 - 1/6)
...................
a(n-3)=1/2(1/(n-3) - 1/(n-1))
a(n-2)=1/2(1/(n-2) - 1/n)
a(n-1)=1/2(1/(n-1) - 1/(n+1))
an=1/2(1/n - 1/(n+2))
则前N项的和为1/2(1 + 1/2 - 1/(n+1) - 1/(n+2) )

Answer 3

an=1/[n(n+2)]=1/(n+2)-1/n
n=(2k-1) an=1/[(2k-1)(2k+1)]=[1/(2k-1)-1/(2k+1)]*(1/2)
n=2k an=1/2k(2k+2)=[1/k-1/(k+1)]*(1/4)
偶数列n=2k
Sn偶数=(1/4)[1-1/(k+1)]
奇数列n=2k-1
Sn奇数=(1/2)*[1-1/(2k+1)]

所以对n=2k时，
Sn=Sn偶数+Sn奇数(2k-1)=3/4-1/(4k+2)-1/(4k+4)
即Sn=(3/4)-1/(2n+2)-1/(2n+4)
n=2k-1时
Sn=Sn偶数(2k-2)+Sn奇数=3/4-1/(4k+2)-1/4k
即Sn=(3/4)-1/(2n+4)-1/(2n+2)
所以对任意n
Sn=(3/4)-1/(2n+2)-1/(2n+4)

Answer 4

an=1/2[1/n-1/(n+2)]
Sn=1/2[1-1/3+1/2-1/4+1/3-1/5+...1/n-1/(n+2)]
=1/2[1+1/2-1/(n+1)-1/(n+2)]
=1/2[3/2-(2n+3)/(n+1)(n+2)]

Answer 5

s=(1/2)[3/2-1/(n+1)-1/(n+2)]