Question

已知数列{a_n}的首项为a₁=1,其前n项和为S_n,且对任意正整数n有n,a_n,S_n成等差数列.
(1)求证:数列{S_n+n+2}成等比数列.
(2)求数列{a_n}的通项公式.

Answer 1

(1)见解析   (2) a_n=2ⁿ-1

(1)因为n,a_n,S_n成等差数列,所以2a_n=S_n+n,由当n≥2时,a_n=S_n-S_n-1,
所以2(S_n-S_n-1)=S_n+n,
即S_n=2S_n-1+n(n≥2),
所以S_n+n+2=2S_n-1+2n+2=2[S_n-1+(n-1)+2].
又S₁+1+2=4≠0,
所以=2,所以数列{S_n+n+2}成等比数列.
(2)由(1)知{S_n+n+2}是以S₁+3=a₁+3=4为首项,2为公比的等比数列,所以S_n+n+2=4×2^n-1=2ⁿ⁺¹,又2a_n=n+S_n,所以2a_n+2=2ⁿ⁺¹,所以a_n=2ⁿ-1.