Question

已知数列{a小n}的首项a1=3,通项a小n=2^np+nq(n属于N,p,q为常数),且a1,a4,a5成等差数列,求

Answer 1

(1)a1=2p+q=3,a4=16p+4q,a5=32p+5q

a1,a4,a5成等差数列,即2(16p+4q)=3+32p+5q

解得p=q=1

(2)an=2^n+n

Sn=西格玛(2^n)+西格玛n=3(1-2^n)/(1-2) + n(n+1)/2=3(2^n-1)+n(n+1)/2