Sn=a^n +k
则,Sn-1=a^(n-1) +k
则an=Sn-Sn-1=a^n-a^(n-1)
=(a-1)a^(n-1)
则a1=(a-1)
所以S1=a +k=a1
则k=-1时,{an}为以(a-1)为首项,a为公比的等比数列.
an= (a-1)a^(n-1)
{bn}满足bn=(cos 0.5nπ)/an,且B2n=b1+b2+...+b2n
则bn=(cos 0.5nπ)/(a-1)a^(n-1)
因为cos 0.5nπ=0,-1,0,1
所以b2,b4...b2n变成以-a2为首项,-a^2为公比的等比数列
则limB2n即求数列的和.=-a2/1-q^2
limB2n=b1+b2+...+b2n=(1-a)a/(1+a^2)