1/[n(n+1)]=(1/n)- [1/(n+1)]。1/[(2n-1)(2n+1)]=1/2[1/(2n-1)-1/(2n+1)]。1/[n(n+1)(n+2)]=1/2{1/[n(n+1)]-1/[(n+1)(n+2)]}。
1数列裂项求和法例题
1/(3n-2)(3n+1)
1/(3n-2)-1/(3n+1)=3/(3n-2)(3n+1)
1/[n(n+1)]=(1/n)- [1/(n+1)]。1/[(2n-1)(2n+1)]=1/2[1/(2n-1)-1/(2n+1)]。1/[n(n+1)(n+2)]=1/2{1/[n(n+1)]-1/[(n+1)(n+2)]}。
1数列裂项求和法例题
1/(3n-2)(3n+1)
1/(3n-2)-1/(3n+1)=3/(3n-2)(3n+1)