等比an>0n=1,a5*a2n-5=2^2n

Question

等比an>0n=1,a5*a2n-5=2^2n

Answer 1

∵a[n]是
.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