已知a_1=4,a_(n+1)=(〖a_n〗^2+4)/(2a_n ),求数列通项公式
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/22 01:32:21
![已知a_1=4,a_(n+1)=(〖a_n〗^2+4)/(2a_n ),求数列通项公式](/uploads/image/z/386665-25-5.jpg?t=%E5%B7%B2%E7%9F%A5a_1%3D4%2Ca_%28n%2B1%29%3D%28%E3%80%96a_n%E3%80%97%5E2%2B4%29%2F%282a_n+%29%2C%E6%B1%82%E6%95%B0%E5%88%97%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
已知a_1=4,a_(n+1)=(〖a_n〗^2+4)/(2a_n ),求数列通项公式
已知a_1=4,a_(n+1)=(〖a_n〗^2+4)/(2a_n ),求数列通项公式
已知a_1=4,a_(n+1)=(〖a_n〗^2+4)/(2a_n ),求数列通项公式
容易发现你这一题是某一年高考题的最后一题的最后一问
a(n+1)=(an²+4)/(2an)
a(n+1) +2=(an²+4an+4)/(2an)=(an +2)²/(2an)
a(n+1)-2=(an²-4an+4)/(2an)=(an -2)²/(2an)
[a(n+1)+2]/[a(n+1)-2]=[(an +2)/(an -2)]²
n=...
全部展开
a(n+1)=(an²+4)/(2an)
a(n+1) +2=(an²+4an+4)/(2an)=(an +2)²/(2an)
a(n+1)-2=(an²-4an+4)/(2an)=(an -2)²/(2an)
[a(n+1)+2]/[a(n+1)-2]=[(an +2)/(an -2)]²
n=1时,(a1+2)/(a1-2)=(4+2)/(4-2)=3>0,假设当n=k(k∈N+)时,(ak+2)/(ak-2)>0
则当n=k+1时,
[a(k+1)+2]/[a(k+1)-2]=[(ak+2)/(ak-2)]²>0
k为任意正整数,因此对于任意正整数n,数列{(an +2)/(an -2)}恒>0
log3[(a(n+1)+2)/(a(n+1)-2)]=log3[(an +2)/(an -2)]²=2log3[(an +2)/(an -2)]
log3[(a(n+1)+2)/(a(n+1)-2)]/log3[(an +2)/(an -2)]=2,为定值
log3[(a1+2)/(a1-2)]=log3(3)=1,数列{log3[(an +2)/(an -2)]}是以1为首项,2为公比的等比数列
log3[(an +2)/(an -2)]=2^(n-1)
(an +2)/(an -2)=3^[2^(n-1)]
{3^[2^(n-1)]-1}an=2×3^[2^(n-1)]+2
an={2×3^[2^(n-1)]+2}/{3^[2^(n-1) -1]}
收起