(1/2)已知数列an满足条件：a1=1.a2=r.(r>0)且{anan+1}是公比为q的等比数列,设bn=a2n-1+a2n,求数列bn的前

an*an+1/an-1*an=q→an+1/an-1=q→an+2/an=q→a(2k+1)=a(2k-1)*q;a(2k)=a(2k-2)*q
a1=1,a2=r→bn=q^(n-1)+r*q^(n-1)→bn=(r+1)*q^(n-1)
∑bn=(r+1)*(q^0+q^1+q^2+...q^(n-1))=(r+1)*(1-q^n)/(1-q)