已知正项等比数列{an}中,对任意的n∈N+,都有lga1+lga2+lga3+……+lgan=n^2+n求{an}的通项公式
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![已知正项等比数列{an}中,对任意的n∈N+,都有lga1+lga2+lga3+……+lgan=n^2+n求{an}的通项公式](/uploads/image/z/15218346-66-6.jpg?t=%E5%B7%B2%E7%9F%A5%E6%AD%A3%E9%A1%B9%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2C%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84n%E2%88%88N%2B%2C%E9%83%BD%E6%9C%89lga1%2Blga2%2Blga3%2B%E2%80%A6%E2%80%A6%2Blgan%3Dn%5E2%2Bn%E6%B1%82%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
已知正项等比数列{an}中,对任意的n∈N+,都有lga1+lga2+lga3+……+lgan=n^2+n求{an}的通项公式
已知正项等比数列{an}中,对任意的n∈N+,都有lga1+lga2+lga3+……+lgan=n^2+n
求{an}的通项公式
已知正项等比数列{an}中,对任意的n∈N+,都有lga1+lga2+lga3+……+lgan=n^2+n求{an}的通项公式
lga1+lga2+...+lgan=lg(a1·a2·a3····an)=n^2+n=>a1·a2·a3····an=e^(n^2+n)
所以a1·a2·a3····a(n-1)= e^((n-1)^2+n-1)
上下两式相除得 an=[e^(n^2+n)]/[e^((n-1)^2+n-1)] =e^(2n)
lga1+lga2+lga3+……+lgan=n^2+n
lg(a1*a2*......an)=lg10^(n^2+n)
a1*a2*......an=10^(n^2+n) ①
所以a1·a2·a3····a(n-1)= 10^((n-1)^2+n-1) ②
①/②an=10^(2n)
令n=1,得lga1=2,即a1=20
lga1+lga2+lga3+……+lgan=n^2+n
lga1+lga2+lga3+……+lgan-1=(n-1)^2+n-1
相减得lgan-lgan-1=2n,即an/an-1=10^(2n)
于是an=100^n an-1=……=100^(n+n-1+n-2+……+2)a1
=2*10^(n^2+n-1)