Question

設數列{}滿足a₁=1,a_n+1=2+1(n∈N^*),則{}的通項公式是(   )

A.2ⁿ-1                   B. 2ⁿ              C. 2ⁿ +1                D.2^n-1

Answer 1

A

解法一:a₁=1,a_n+1=22ⁿ +1a₂=3,排除B、C、D.

解法二:∵a_n+1=2a_n+1a_n+1+1=2(a_n+1),

∵a_n+1≠0,∴{a_n+1}是以a₁+1=2為首項,公比為2的等比數列.

∴a_n+1=(a₁+1)·2^n-1=2ⁿa_n=2ⁿ-1.