已知数列{an}满足a1=4,an=4-4/(an-1)(n大于等于2)求数列an的通项公式.
热心网友
an=4-4/(an-1)(n大于等于2)an=[4(an-1)-4]/(an-1)an-2=[4(an-1)-4]/(an-1)-2=2*[(an-1)-2]/(an-1)1/(an-2)=(an-1)/{2*[(an-1)-2]}=1/2+1/[(an-1)-2]设{bn}={1/(an-2)},则bn是等差函数。公差是1/2求得bn=1/2*(n-1)+b1,b1=1/(a1-2)=1/2,bn=n/2带入{bn}={1/(an-2)}式中即可得an=2+2/n
热心网友
an=4-4/(an-1)(n大于等于2)an=[4(an-1)-4]/(an-1)an-2=[4(an-1)-4]/(an-1)-2=2*[(an-1)-2]/(an-1)1/(an-2)=(an-1)/{2*[(an-1)-2]}=1/2+1/[(an-1)-2]设{bn}={1/(an-2)},则bn是等差函数。公差是1/2求得bn=1/2*(n-1)+b1,b1=1/(a1-2)=1/2,bn=n/2带入{bn}={1/(an-2)}式中即可得an=2+2/n