∵an的表达式为一次函数,∴{an}是等差数列an=a1+(n-1)d=dn+(a1-d)=4n-5/2∴d=4,a1=3/2Sn=d/2*n²+(a1-d/2)n=an²+bn∴a=2,b=-1/2∴lim(n→∞)[2^n-(-1/2)^n]/[2^n+(-1/2)^n]=lim(n→∞)[1-(-1/4)^n]/[1+(-1/4)^n]=(1-0)/(1+0)=1
