Question

已知数列{a n }中,a 1 =1,前n项和为S n 且S n+1 = S n +1,n∈N * .(1)求数列{a n }的通项公式.(2

已知数列{a n }中,a 1 =1,前n项和为S n 且S n+1 = S n +1,n∈N * .(1)求数列{a n }的通项公式.(2)求数列{ }的前n项和T n .

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Answer 1

(1) a n =( ) n-1    (2) T n =3[1-( ) n ]

(1)由S n+1 = S n +1, 得当n≥2时S n = S n-1 +1, ∴S n+1 -S n = (S n -S n-1 ), 即a n+1 = a n ,∴ = , 又a 1 =1,得S 2 = a 1 +1=a 1 +a 2 ,∴a 2 = , ∴ = , ∴数列{a n }是首项为1,公比为 的等比数列, ∴a n =( ) n-1 . (2)∵数列{a n }是首项为1,公比为 的等比数列, ∴数列{ }是首项为1,公比为 的等比数列, ∴T n = =3[1-( ) n ].