求和:1:Sn=1/1*3+1/2*4+1/3*5+.+1/n*(n+2) 2:Sn=1/1*2+1/2*3+1/3*4+.+1/n(n+1)

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求和:1:Sn=1/1*3+1/2*4+1/3*5+.+1/n*(n+2) 2:Sn=1/1*2+1/2*3+1/3*4+.+1/n(n+1)

求和:1:Sn=1/1*3+1/2*4+1/3*5+.+1/n*(n+2) 2:Sn=1/1*2+1/2*3+1/3*4+.+1/n(n+1)
求和:1:Sn=1/1*3+1/2*4+1/3*5+.+1/n*(n+2) 2:Sn=1/1*2+1/2*3+1/3*4+.+1/n(n+1)

求和:1:Sn=1/1*3+1/2*4+1/3*5+.+1/n*(n+2) 2:Sn=1/1*2+1/2*3+1/3*4+.+1/n(n+1)
1:Sn=1/1*3+1/2*4+1/3*5+.+1/n*(n+2)
= 1/2*(1-1/3) + 1/2*(1/2 - 1/4) + ...+ 1/2*(1/n - 1/(n+2))
= 1/2 ( 1+ 1/2 - 1/(n+1) - 1/(n+2))
2:Sn=1/1*2+1/2*3+1/3*4+.+1/n(n+1)
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/n - 1/(n+1)
= 1- 1/(n+1)
= n / (n+1)

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

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

收起

S2=1-1/2+1/2-1/3...............+1/n-1/(n+1)=n/(n+1)

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