a(n+1)^2/an^2=(a1q^n)^2/[a1q^(n-1)]^2=[q^n/q^(n-1)]^2=q^2=(√2/2)^2=1/2数列{an^2}是以2为首项,1/2为公比的等比数列,令q'=1/2.a1^2+a2^2+...+a10^2=(a1^2)(1-q'^10)/(1-q')=4(1-1/1024)/(1-1/2)=8(1-1/1024)=102...