数列{an}对任意n∈N*满足an+1=an+a2,且a3=6,则a10等于A.24B.27C.30D.32
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B
解析分析:根据题意,an+1=an+a2,且a3=6,所以a2=3.a4=6+3=9,a5=9+3=12,以此类推,能够求出a10.
解答:∵an+1=an+a2,且a3=6,∴a3=2a2,∴a2=3.∴a4=6+3=9,a5=9+3=12,a6=12+3=15,a7=15+3=18,a8=18+3=21,a9=21+3=24,a10=24+3=27.故选B.
点评:本题考查数列的递推公式,解题时要注意递推思想的合理运用.