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1. 已知[Sk=1k+1+1k+2+1k+3+?+12k][(k=][1,2,3,?)],则[Sk+1]等于( )
A. [Sk+12(k+1)] B. [Sk+12(k+1)-1k+1]
C. [Sk+12k+1-12k+2] D. [Sk+12k+1+12k+2]
A. [Sk+12(k+1)] B. [Sk+12(k+1)-1k+1]
C. [Sk+12k+1-12k+2] D. [Sk+12k+1+12k+2]