Question

设a i 大于0，i =1,2,3...,k,试求数列极限n 次根号下a1n次方+a2n次方+……+

接着上面a k n次方

Answer 1

利用极限的夹逼性，进行适当放缩，即可破题，下面是具体解题步骤

Answer 2

解: ∵a[n]是等比数列。又有a[5]*a[2n-5]=2^2n ∴a[1]*a[2n-1]=2^2n
设a[n]得公比为q. ∵a[1]*q^n-1=a[n]， a[2n-1]=a[1]*q^2n-2，
∴a[1]*a[2n-1]=(a[1]*q*n-1)²=a[n]² ∴a[n]²=2^2n 从而 a[n]=2^n
∵log2 a[1]+log2 a[2]+…+log2a[2n-1]=log2(a[1]*a[2]*…a[2n-1])
=log2[(a[1]*a[2n-1])*(a[2]*a[2n-2])…
=log2[2^n²]
=n^2