Question

數(shù)列{a_n}的前n項(xiàng)和記為S_n,a₁=t,點(diǎn)(S_n,a_n+1)在直線y=2x+1上,n∈N^*,若數(shù)列{a_n}是等比數(shù)列,則實(shí)數(shù)t=　　　.

Answer 1

1

【思路點(diǎn)撥】得出關(guān)于a_n+1,S_n的式子,降低一個(gè)角標(biāo)再得一個(gè)關(guān)于a_n,S_n-1的式子,兩個(gè)式子相減后得出a_n+1,a_n的關(guān)系,可得數(shù)列{a_n}中,a₂,a₃,a₄,…為等比數(shù)列,只要等于上面數(shù)列的公比即可.
解：由題意得a_n+1=2S_n+1,
a_n=2S_n-1+1(n≥2),
兩式相減得a_n+1-a_n=2a_n,即a_n+1=3a_n(n≥2),
所以當(dāng)n≥2時(shí),{a_n}是等比數(shù)列,
要使n≥1時(shí),{a_n}是等比數(shù)列,則只需
==3,從而t=1.