正项数列{an},a1=1,2 an²=an+1²+an-1²,n≥2,求a6=?
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![正项数列{an},a1=1,2 an²=an+1²+an-1²,n≥2,求a6=?](/uploads/image/z/5456562-42-2.jpg?t=%E6%AD%A3%E9%A1%B9%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%2Ca1%3D1%2C2+an%26%23178%3B%3Dan%2B1%26%23178%3B%2Ban-1%26%23178%3B%2Cn%E2%89%A52%2C%E6%B1%82a6%3D%3F)
正项数列{an},a1=1,2 an²=an+1²+an-1²,n≥2,求a6=?
正项数列{an},a1=1,2 an²=an+1²+an-1²,n≥2,求a6=?
正项数列{an},a1=1,2 an²=an+1²+an-1²,n≥2,求a6=?
由2 an²=an+1²+an-1²得 :an+1² - an² = an² - an-1²
an²是等差数列,设该数列公差为 q
则 a6² = a1² + (6-1)q =1+5q
例如,如果 a2=2 ,则 q=3,a6= 4
如果 a2=3 ,则 q=8,a6= 根号41