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bn2 1 bn2-1 1 4n2- 1 = 1 2 ( 1 2n-1 - 1 2n+1 ) 解答:解:(1)∵当n≥2sn=1- (2n-1) 2 (sn-sn-1) ∴(2n+1)sn=(2n-3)sn即bn=bn-1+2,, 又∵b1=3×s1=3× 2 3 =2 ∴bn=2+2(n-1)=2n (2)∵ 1 bn2
Fundholding 'may save pounds 1bn a year' 来自 EBSCO 喜欢 0 阅读量: 24 摘要: The article focuses on a report by the think tank Policy Exchange that savings of at least pounds 1 billion per year will be delivered by implementing general practitioner (GP) fundholding in Great Britain. It...
如图.点M.N分别是正五边形ABCDE的边BC.CD上的点.且BM=CN.AM交BN于点P. (1)求证:△ABM≌△BCN, (2)求∠APN的度数.