数列a1=1a2=0.5 an(an-1+an+1)=2(an+1)(an-1),求数列2011项

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数列a1=1a2=0.5an(an-1+an+1)=2(an+1)(an-1),求数列2011项数列a1=1a2=0.5an(an-1+an+1)=2(an+1)(an-1),求数列2011项数列a1

数列a1=1a2=0.5 an(an-1+an+1)=2(an+1)(an-1),求数列2011项
数列a1=1a2=0.5 an(an-1+an+1)=2(an+1)(an-1),求数列2011项

数列a1=1a2=0.5 an(an-1+an+1)=2(an+1)(an-1),求数列2011项
a[n](a[n-1]+a[n+1])=2a[n+1]*a[n-1]
我用方括号表示下角标.
把此等式展开,两边同除以a[n-1]*a[n]*a[n+1]
那么得到1/a[n+1] + 1/a[n-1] = 2/a[n]
1/a[n+1]-1/a[n]=1/a[n]-1/a[n-1]
所以可以看出1/a[n]是等差数列.
1/a1=1,1/a2=2
所以公差为1,1/a[n]=n
a[n]=1/n
a2011 = 1/2011

an(an-1+an+1)=2(an+1)(an-1),(an+1)*(an-an-1)=(an-1)(an+1-an),(an/an-1)-1=1-an/an+1;令an/an+1=bn;