求数列1/2^2+4,1/4^2+8,1/6^2+12.1/(2n)^2+4n的前n项和和sn
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/09/23 04:23:03
求数列1/2^2+4,1/4^2+8,1/6^2+12.1/(2n)^2+4n的前n项和和sn
1/{2^2+4]+ 1[/4^2+8]+1/[6^2+12]+...+1/[(2n)^2+4n]
consider
1/[(2n)^2+4n]
=1/[4n(n+1)]
=(1/4)[1/n -1/(n+1)]
1/{2^2+4]+ 1[/4^2+8]+1/[6^2+12]+...+1/[(2n)^2+4n]
=(1/4) [ (1/1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1)]
=(1/4)( 1-1/(n+1) )
consider
1/[(2n)^2+4n]
=1/[4n(n+1)]
=(1/4)[1/n -1/(n+1)]
1/{2^2+4]+ 1[/4^2+8]+1/[6^2+12]+...+1/[(2n)^2+4n]
=(1/4) [ (1/1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1)]
=(1/4)( 1-1/(n+1) )
求数列{(2n-1)*1/4的n次方}的前n项和Sn
求数列4,9,16,.,3n-1+2^n,.前n项的和Sn
已知数列{an}的前n项和为Sn=1+2+3+4+…+n,求f(n)= Sn /(n+32)Sn+1的最大值
在数列{an}中,a1=2,sn=4A(n+1) +1 ,n属于N*.求数列{an}的前n项和Sn
求数列1/2^2+4,1/4^2+8,1/6^2+12.1/(2n)^2+4n的前n项和和sn
设数列{an}的前n项和Sn=(-1)^n(2n^2+4n+1)-1
设数列{an}的前n项和Sn=(-1)^n(2n^2+4n+1)-1,
已知an=5n(n+1)(n+2)(n+3),求数列{an}的前n项和Sn
已知数列{An}的前n项和为Sn,A2n=n+1(n∈N*),S2n-1=4n^2-2n+1(n∈N*),求数列{An}
数列:已知数列{an}前 n项和为Sn,且a1=2,4Sn=ana(n+1).求数列{an}的通项公式.
已知数列an的前n项和Sn,求数列的通项公式.(1)Sn=3n²-n (2)Sn=2n+1
已知数列{an}的前n项和Sn=1/3n(n+1)(n+2),试求数列(1/an)的前n项和