高中数学数列问题

显示全部楼层 · 2009-8-16 12:37:50

1.an=1/[(1+n)*n/2]so an=2/n*(n+1)=2*(1/n-1/(n+1)Sn=2(1-1/2+1/2-1/3+1/3-1/4........+1/n-1/(n+1)]右边中间相都抵消得到Sn=2*(1-1/(n+1））=2n/(n+1)2.f(x)+f(1-x)=1/(2^x+根号2)+1/[2^(1-x)+根号2]=1/(2^x+根号2)+2^x/[2+根号2*2^x]=(1+2^x/根号2)*[1/(2^x+根号2)]=[(2^x+根号2)/根号2]*[1/(2^x+根号2)]=1/根号2=根号2/2f(-5)+f(-4)+...+f(0)+...+f(5)+f(6)=根号...

千问 · 2009-8-16 12:37:50

1.An=2/[n(n+1)]=2[1/n-1/(n+1)]Sn=1/1 + 1/(1+2) +1/(1+2+3)+..+1/(1+2+...+n)=2[1-1/2 +1/2-1/3 +1/3-1/4+...+1/n-1/(n+1)]=2[1-1/(n+1)]=2n/(n+1)2.f(x)+f(1-x)=1/(2^x+...

千问 · 2009-8-16 12:37:50

1.An=1/（1+2+3+4+...n)=2/[(n+1)n]=2*[1/n-1/(n+1)];sn=2*[(1-1/2)+(1/2-1/3)+....+(1/n-1/(n+1))]=2*[1-1/(n+1)]=2n/(n+1);2....

千问 · 2009-8-16 12:37:50

1.通项是：An = 2/(n^2 + n) = 2*(1/n - 1/(n+1));如此裂项相加可得：Sn = 2-2/(n+1)...

千问 · 2009-8-16 12:37:50

1+2+3+4+...n=(n+1)*n/2An=2/(n+1)*n=2*(1/n-1/(n+1))Sn=2*(1-1/2+1/2-1/3+...+1/(1+n))=2*(1-1/(n+1))...