高中數(shù)學(xué)北師大版必修5前n項和sn的求法導(dǎo)學(xué)案(更新版)
【正文】 n+1n2. :(1)當(dāng) n為奇數(shù)時 , Sn=(a1+a3+a5+? +an)+(a2+a4+a6+? +an1) = + = 2 2+33 n+2n. (2)設(shè) Tn=03 n+1, ∴T n= + = , ∴S n=Tn+(2+22+? +2n) = . 應(yīng)用三 :(1) = ( ), ∴S n= (1 + + +?? + + )= (1+ ) = . (2) = ( ) ∴S n= [( )+( )+( )+? +( )]= ( )= . 基礎(chǔ)智能檢測 ∵a n=(1)n(3n2),∴a 1+a2+? +a10=1+47+10? 25+28=(1+4)+(7+10)+? +(25+28)=35=15. a2=2a1=2,a3=a2+1=3,a4=2a3=6,a5=a4+1=7,a6=2a5=14,所以 S6=1+2+3+6+7+14=33. +12 ∵a n+1an=2n,∴a n=(anan1)+(an1an2)+? +(a2a1)+a1=2n1+2n2+? +22+2+2= +2=2n2+2=2n,∴Sn= =2n+12. :∵a n= = =2( ), ∴S n=2[(1 )+( )+? +( )]=2(1 )= . 全新視角拓展 +3242+? +(1)n+1n2=(1)n+1 設(shè)等式右邊的數(shù)的絕對值構(gòu)成數(shù)列 {an},∵a 2a1=2,a3a2=3,a4a3=4,?, anan1=n,以上所有等式相加可得 ana1=2+3+4+? +n,即 an=1+2+3+? +n= ,再觀察各式的符號可知第 n 個等式為 :1222+3242+? +(1)n+1n2=(1)n+1 . :(1)由 (2n1)an2n=0,得 (an2n)(an+1)=0. 由于 {an}是正項數(shù)列 ,所以 an=2n. (2)由 an=2n,bn= , 則 bn= = ( ), Tn= (1 + +? + + )= (1 )= .
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