数学数列的应用题
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数学数列的应用题
2Sn=an^2+an
2S(n-1)=a(n-1)^2+a(n-1)
2an=an^2+an-a(n-1)^2-a(n-1)
an^2-a(n-1)^2-[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-1]=0
an-a(n-1)-1=0
an-a(n-1)=1
{an}是等差数列
s1=a1,2a1=a1^2+a1
a1=1
an=a1+(n-1)=1+n-1=n
an=n
bn=1/ana(n+1)=1/n(n+1)=[1/n-1/(n+1)]
Tn=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+1/n-1/(n+1)=1-1/(n+1)
2S(n-1)=a(n-1)^2+a(n-1)
2an=an^2+an-a(n-1)^2-a(n-1)
an^2-a(n-1)^2-[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-1]=0
an-a(n-1)-1=0
an-a(n-1)=1
{an}是等差数列
s1=a1,2a1=a1^2+a1
a1=1
an=a1+(n-1)=1+n-1=n
an=n
bn=1/ana(n+1)=1/n(n+1)=[1/n-1/(n+1)]
Tn=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+1/n-1/(n+1)=1-1/(n+1)