证明:1/(1×3)+1/(2×4)+1/(3×5)+...+1/[n(n+2)]=3/4-(2n+3)/[2(n+1)
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证明:1/(1×3)+1/(2×4)+1/(3×5)+...+1/[n(n+2)]=3/4-(2n+3)/[2(n+1)(n+2)]
我能算到3/4-1/2(n+1)-1/2(n+2),可是就是不知道怎样得到3/4-(2n+3)/[2(n+1)(n+2)]
我能算到3/4-1/2(n+1)-1/2(n+2),可是就是不知道怎样得到3/4-(2n+3)/[2(n+1)(n+2)]
![证明:1/(1×3)+1/(2×4)+1/(3×5)+...+1/[n(n+2)]=3/4-(2n+3)/[2(n+1)](/uploads/image/z/17974717-61-7.jpg?t=%E8%AF%81%E6%98%8E%EF%BC%9A1%2F%EF%BC%881%C3%973%29%2B1%2F%282%C3%974%29%2B1%2F%283%C3%975%29%2B...%2B1%2F%5Bn%28n%2B2%29%5D%3D3%2F4-%282n%2B3%29%2F%5B2%28n%2B1%29)
这是数学的一种方法,叫做“裂项相消法”,常用于分式的相加或相乘中,可以写 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/-1/)+(1/n+1/)]/2
=(1+1/2-1/-1/)/2
=3/4-1/2(n+1)-1/2(n+2)
=3/4-(2n+3)/2(n+1)(n+2
另外,再教你个类似的1/n(n-a)=(1/(n-a)-1/n)/a,这是解决此类方法的通式,希望回答可以帮到你!
=[(1-1/3)+(1/2-1/4)+(1/3-1/5).+(1/-1/)+(1/n+1/)]/2
=(1+1/2-1/-1/)/2
=3/4-1/2(n+1)-1/2(n+2)
=3/4-(2n+3)/2(n+1)(n+2
另外,再教你个类似的1/n(n-a)=(1/(n-a)-1/n)/a,这是解决此类方法的通式,希望回答可以帮到你!
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