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求数学,若数列{An}满足An=2n%1/3的n次方,求{An}的前n项和sn

sn=2n的前n项和-1/3的n次方的前n项和 =n(n+1)-(1-(1/3)^n)/(3*(1-1/3)) =n(n+1)+0.5(1/3)^n-0.5

n=1,a1=s1=121=2,n≥2,an=snsn1.看不懂再问

用错位相减的方法求3an=(2n-1)*3^(n+1)3sn-sn 相减就可以了.

a1=3a2=3*3^2a3=5*3^3Sn=3+3*3^2+5*3^3++(2n-1)*3^n3Sn=3^2+3*3^3+5*3^4++(2n-1)*3^(n+1)3Sn-Sn=3^2+3*3^3+5*3^4++(2n-1)*3^(n+1)-[3+3*3^2+5*3^3++(2n-1)*3^n]2Sn=(2n-1)*3^(n+1)+3-2*[3+3^2+3^3++3

解:sn=a1+a2++an =(2^1+2*1-1)+(2^2+2*2-1)++(2^n+2n-1) =(2^1+2^2++2^n)+(2*1+2*2++2*n)+(1+1++1) =2(1-2^n)/(1-2)+2n(1+n)/2+n =2^(n+1)-2+n^2+n+n =2^(n+1)+n^2+2n-2

解:bn=an+3=2n+1+3 Sn=b1+b2++bn=[3+5++(2n+1)]+(3+3++3)=[1+3+5++(2n+1)]-1+3(3-1)/(3-1)=(n+1)+3 +

an=(2n-1)3^(n-1) = 2[ n.3^(n-1) ] - 3^(n-1)Sn = a1+a2++an = 2[∑(i:1->n) i.3^(i-1) ] - (3^n-1)/2letS = 1.3^0+ 2.3^1+.+n.3^(n-1) (1)3S = 1.3^1+ 2.3^2+.+n.3^n (2)(2)-(1)2S = n.3^n - [ 3^0+3^1+.+3^(n-1) ] =n.2^n - (3^n-1)/2S = 2n.2^n - (3^n-1)Sn

an=a^2n-a^nan=a^n*a^n-a^nan=a^n(a^n-1)两边除以a^na^n-1=1/n两边除以a^(n-1)an=1/n-1an=(1-n^2)/n则a1=0,a2=-3/2,a3=-8/3,a4=-15/4Sn:公式忘了

设An = (2n-1)*3^n=2n*3^n-3^n=an+bnan=2n*3^n;bn=-3^n前面一个是等差数列和等比数列的积,用错位相减求和,后面的是Q=3的等比数列求和两个和做差即可Sn1=2*3^1

anbn=(2n+1)3*n,前n项和就为Tn 所以Tn=3x3+5x3*2+7x3*3++(2n+1)3*n .(1) 3Tn= 3x3*2+5x3*3++(2n-1)3*n+(2n+1)3*(n+1) .(2) (1)-(2)得: -2Tn=3x3+2x3*2+2x3*3++2x3*n-(2n+1)3*(n+1) =9+18(1-3*n)/(1-3)-(2n+1)3*(n+1) =9-9(1-3*n)-(2n+1)3*(n+1) 所以Tn=[9-9(1-3*n)-(2n+1)3*(n+1)]/(-2) 楼主再约分即可 望采纳,谢谢

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