3个回答
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题目里没有x、f(x)啊;
假如f(x)=1/(1*3)+1/(2*4)+1/(3*5)+...+1/[x(x+2)],x属于N;
x=6,
f(x)
=f(6)
=1/(1*3)+1/(2*4)+1/(3*5)+...+1/[6(6+2)],
=1/3+1/8+1/15+1/24+1/35+1/48
=560/1680+210/1680+112/1680+70/1680+48/1680+35/1680
=965/1680
=193/336
假如f(x)=1/(1*3)+1/(2*4)+1/(3*5)+...+1/[x(x+2)],x属于N;
x=6,
f(x)
=f(6)
=1/(1*3)+1/(2*4)+1/(3*5)+...+1/[6(6+2)],
=1/3+1/8+1/15+1/24+1/35+1/48
=560/1680+210/1680+112/1680+70/1680+48/1680+35/1680
=965/1680
=193/336
追问
那个x写的有点丑 跟n一样 能给出最终化简的含x的式子吗
追答
上面有错误:
f(6)
=1/(1*3)+1/(2*4)+1/(3*5)+...+1/[6(6+2)],
=1/3+1/8+1/15+1/24+1/35+1/48
=560/1680+210/1680+112/1680+70/1680+48/1680+35/1680
=1035/1680
=69/112
1/x-1/(x+2)=2/[x(x+2)]
1/[x(x+2)]=(1/2)[1/x-1/(x+2)]
f(x)=1/(1*3)+1/(2*4)+1/(3*5)+...+1/[x(x+2)],
=(1/2)[1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+...+1/x-1/(x+2)]
=(1/2)[1+1/2+...-1/(x+1)-1/(x+2)]
=3/4-(2x+3)/[2(x+1)(x+2)]
f(6)=3/4-(12+3)/[2(6+1)(6+2)]=3/4-15/112=69/112
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2014-10-04
展开全部
稳定化雾凇沆砀
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