K= -2n/3 -1 1/K1K2+1/K2K3+.1/K(n-1)Kn=?
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Kn= -2n/3 -1=(-2n-3)/3,K(n-1)=(-2n-1)/3;
1/Kn=-3/(2n+3),1/K(n-1)=-3/(2n+1);
1/K(n-1)Kn=9 / (2n+1)(2n+3)
1/K(n-1) - 1/Kn
=-3[1/(2n+1)-1/(2n+3)]
=-3* 2/ (2n+1)(2n+3)
=-6/ (2n+1)(2n+3)
所以,1/K(n-1)Kn=-3/2[1/K(n-1) - 1/Kn]
因此,
1/K1K2+1/K2K3+.1/K(n-1)Kn (n=2,3,4,……,n)
=-3/2[(1/K1-1/K2)+(1/K2-1/K3)+……+(1/K(n-1) - 1/Kn)]
=-3/2[1/K1 - 1/Kn]
=-3/2[-3/5 - (-3)/(2n+3)]
=……最后就辛苦下你自己了哈!
1/Kn=-3/(2n+3),1/K(n-1)=-3/(2n+1);
1/K(n-1)Kn=9 / (2n+1)(2n+3)
1/K(n-1) - 1/Kn
=-3[1/(2n+1)-1/(2n+3)]
=-3* 2/ (2n+1)(2n+3)
=-6/ (2n+1)(2n+3)
所以,1/K(n-1)Kn=-3/2[1/K(n-1) - 1/Kn]
因此,
1/K1K2+1/K2K3+.1/K(n-1)Kn (n=2,3,4,……,n)
=-3/2[(1/K1-1/K2)+(1/K2-1/K3)+……+(1/K(n-1) - 1/Kn)]
=-3/2[1/K1 - 1/Kn]
=-3/2[-3/5 - (-3)/(2n+3)]
=……最后就辛苦下你自己了哈!
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