若数列{an}满足a1=5,an+1={[a(n+1)^2]/2an}+an/2 (n∈N+),则其{an}的前10项和为
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![若数列{an}满足a1=5,an+1={[a(n+1)^2]/2an}+an/2 (n∈N+),则其{an}的前10项和为](/uploads/image/z/15255014-14-4.jpg?t=%E8%8B%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E6%BB%A1%E8%B6%B3a1%3D5%2Can%2B1%3D%7B%5Ba%28n%2B1%29%5E2%5D%2F2an%7D%2Ban%2F2+%28n%E2%88%88N%2B%29%2C%E5%88%99%E5%85%B6%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8D10%E9%A1%B9%E5%92%8C%E4%B8%BA)
若数列{an}满足a1=5,an+1={[a(n+1)^2]/2an}+an/2 (n∈N+),则其{an}的前10项和为
若数列{an}满足a1=5,an+1={[a(n+1)^2]/2an}+an/2 (n∈N+),则其{an}的前10项和为
若数列{an}满足a1=5,an+1={[a(n+1)^2]/2an}+an/2 (n∈N+),则其{an}的前10项和为
等式两边同乘以2an,变为:2an*a(n+1)=[a(n+1)]^2*an^2
即[a(n+1)]^2-2an*[a(n+1)]+an^2=0
所以a(n+1)=an
即an=a1=5
前十项和就是50.
P.S不知道你题目是不是这样的,看着有点乱,不是的话,明天再帮你看看.
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