在数列{an}中,an=4n-5/2,an=4n-5/2,a1+a2+...+an=an^2+bn,其中n属于N*,a、b为常数,求lim在数列{an}中,an=4n-（5/2）,an=4n-5/2,a1+a2+...+an=an^2+bn,其中n属于N*,a、b为常数,求lim[(a^n-b^n)/(a^n+b^n)]

a1+a2+...+an=a*n^2+bn
an=4n-5/2，易知｛an｝为等差数列
利用等差数列求和公式得：
n[3/2+4n-（5/2）]/2=a*n^2+bn
n(4n-1)=2a*n^2+2bn
4n^2-n=2a*n^2+2bn
2a=4,2b=-1
a=2,b=-1/2
lim(n--> +∞）[(a^n-b^n)/(a^n+b^n)]
=lim(n--> +∞）[(2^n-(-1/2)^2]/[^n+(-1/2)^n]
=lim(n--> +∞）[1-(-1/4)^2]/[+(-1/4)^n] （上下同时处以2^n)
=(1+0)/(1+0)
=1

a1=4×1-5/2=3/2
a2=4×2-5/2=11/2
a1=a+b=3/2
a3=4a+2b=11/2
解得
a=5/4 b=1/4 a/b=5
lim[(a^n-b^n)/(a^n+b^n)]
=lim[(a/b)^n-1]/[(a/b)^n+1]
=lim[(5^n-1)/(5^n+1)]
=lim[(1-1/5^n)/(1+1/5^n)]
n->+∞时，1/5^n->0 lim[(a^n-b^n)/(a^n+b^n)]=1
n->0时，1/5^n->0 1-1/5^n->0 lim[(a^n-b^n)/(a^n+b^n)]=0

a1+a2+...+an=a*n^2+bnan=4n-5/2,易知｛an｝为等差数列利用等差数列求和公式得：n[3/2+4n-（5/2）]/2=a*n^2+bnn(4n-1)=2a*n^2+2bn4n^2-n=2a*n^2+2bn2a=4,2b=-1a=2,b=-1/2lim(n--> +∞）[(a^n-b^n)/(a^n+b^n)]=lim(n-...