设Sn为数列{an}的前n项和,Sn=(-1)^n an - 1/(2^n),n∈N*,则 (1)a3=___ (2)S1+S2+…+S100=___
问题描述:
设Sn为数列{an}的前n项和,Sn=(-1)^n an - 1/(2^n),n∈N*,则 (1)a3=___ (2)S1+S2+…+S100=___
答
Sn=(-1)^n*an-1/2^nS(n-1)=(-1)^(n-1)*a(n-1)-1/[2^(n-1)]两式相减得:an=(-1)^n*an-(-1)^(n-1)*a(n-1)+1/2^n.①令n=4a4=a4-a3+1/16==>a3= - 1/16(2)在①中令n=2ka(2k)=a(2k)+a(2k-1)+1/[2^(2k)]a(2k-1)= -1/[4^(2k)...