判断级数n*cos^2(nπ/3)/2^n_百度作业帮
判断级数n*cos^2(nπ/3)/2^n
一朵浮云6609
∵数列{a[n]}的通项a[n]=n^2[(cosnπ/3)^2-(sinnπ/3)^2],前n项和为S[n]∴a[n]=n^2Cos(2nπ/3)∴S[n]=1^2(-1/2)+2^2(-1/2)+3^2+4^2(-1/2)+5^2(-1/2)+6^2+...+n^2Cos(2nπ/3)当n=3k-2,即:k=(n+2)/3时:S[3k-2]=(-1/2)[1^2+2^2+...+(3k-2)^2]+3^3/2[1^2+2^2+...+(k-1)^2]=-(3k-2)(3k-1)(6k-3)/12+27(k-1)k(2k-1)/12=[-n(n+1)(2n+1)+(n-1)(n+2)(2n+1)]/12=(2n+1)(-n^2-n+n^2+n-2)/12=-(2n+1)/6当n=3k-1,即:k=(n+1)/3时:S[3k-1]=(-1/2)[1^2+2^2+...+(3k-1)^2]+3^3/2[1^2+2^2+...+(k-1)^2]=-(3k-1)3k(6k-1)/12+27(k-1)k(2k-1)/12=[-n(n+1)(2n+1)+(n-2)(n+1)(2n-1)]/12=(n+1)(-2n^2-n+2n^2-5n+2)/12=(n+1)(-6n+2)/12=-(n+1)(3n-1)/6当n=3k,即:k=n/3时:S[3k]=(-1/2)[1^2+2^2+...+(3k)^2]+3^3/2(1^2+2^2+...+k^2)=-3k(3k+1)(6k+1)/12+27k(k+1)(2k+1)/12=[-n(n+1)(2n+1)+n(n+3)(2n+3)]/12=n(-2n^2-3n-1+2n^2+9n+9)/12=n(6n+8)/12=n(3n+4)/6
其他类似问题
扫描下载二维码求幂级数∑(n=1,∞) x^n/n·3^n的收敛域 求过程_百度知道
求幂级数∑(n=1,∞) x^n/n·3^n的收敛域 求过程
提问者采纳
发散当x=-3时;3;n·3^n)=1/: lim(1/(1/,为调和级数已经做过。为收敛的交错级数收敛域为[-3;[(n+1)3^(n+1)]/,故收敛半径为3当x=3时
提问者评价
按照你说的,真的成功了,好开心,谢谢你!
其他类似问题
为您推荐:
其他2条回答
3收敛半径R=3,当X=3是级数发散,X=-3是级数收敛,收敛域[-3求收敛域只比系数系数:LIM|a(n+1)/a(n)|=1/
∑(n=1,∞) x^n/n·3^n化简以后=∑(n=1,∞)1/n·∑(n=1,∞)x^n/3^n=0··∑(n=1,∞)(x/3)^n=0
呀!我也忘了,你看看数学分析,一下就会了,我这么做是绝对收敛了。不对呀!
参考资料:
等待您来回答
下载知道APP
随时随地咨询
出门在外也不愁数列{an}的通项an=n^2(cos^2(nπ)/3_百度知道
数列{an}的通项an=n^2(cos^2(nπ)/3
url=ocIR1oBS9ZTMTkgVlsuKlwJgJmu69GK9RbwCvk3lEZB0SCz7u0RCWNfUwZL9MgHl8nsMwmrRKm9V9f0enQ0_Kq" target="_blank">http?url=ocIR1oBS9ZTMTkgVlsuKlwJgJmu69GK9RbwCvk3lEZB0SCz7u0RCWNfUwZL9MgHl8nsMwmrRKm9V9f0enQ0_Kq可以给我一份详细答案吗://zhidao:////link.baidu.baidu<a href="http
提问者采纳
3时.;4^n =[16-(9n+16)/4 =(1/.+n/12 =[-n(n+1)(2n+1)+n(n+3)(2n+3)]/12 =(2n+1)(-n^2-n+n^2+n-2)/,即;3时.;2[1^2+2^2+;12 =-(n+1)(3n-1)/(n4^n)
∴b[n]=(9n+4)/12+27k(k+1)(2k+1)/12 =[-n(n+1)(2n+1)+(n-1)(n+2)(2n+1)]/..: S[3k] =(-1/4^n)+2(1/.+1/2)(n/(1-1/4^n+2/2)[1^2+2^2+;4^n] =(9/4^(n+1) =[4-(3n+4)/4^1]+[(9/,即.;4)(1-1/12 =[-n(n+1)(2n+1)+(n-2)(n+1)(2n-1)]/.+k^2) =-3k(3k+1)(6k+1)/12 =-(2n+1)/6
(2)∵当n=3k时.;2)(2/4^4+;2)+3^2+4^2(-1/(2*4^n)=(9/4^(n+1) =(1/4^(n+1) ∴3R[n]/4^n)+2/2+(2/.+n/..+(k-1)^2] =-(3k-2)(3k-1)(6k-3)/4^n
设R[n]=1/9
∴T[n] =[(9/4) =[4-(3n+4)/.+(k-1)^2] =-(3k-1)3k(6k-1)/12 =(n+1)(-6n+2)/..;4^1+2/12 ∴R[n]=[4-(3n+4)/3)
当n=3k-2;6=k(9k+4)/:k=n/12 =(n+1)(-2n^2-n+2n^2-5n+2)/4^n) =2-(3n/2+2)/..;4^2+3/4^1+1/4)-n/4^n) =(9/4^3+.+1/,即;4^2]+.;4^3+3/12+27(k-1)k(2k-1)/6
当n=3k:k=(n+2)/4^n)-n/..+n^2Cos(2nπ/12+27(k-1)k(2k-1)/4^1+2/4)(1-1/3)^2-(sinnπ/..;2 ∵b[n]=S[3n]/4^(n+1) =(1/:k=(n+1)/4^1+1/(1-1/12 =n(3n+4)/6=3k(9k+4)/.;4^2)+2/2)(1/4^n 则R[n]/2)+6^2+;2(1^2+2^2+;4^2+1/,前n项和为S[n] ∴a[n]=n^2Cos(2nπ/12 =n(-2n^2-3n-1+2n^2+9n+9)/4^n)+2/.+n/.;3)/.;4^n)/4^n)/4=1/3)(1-1/3时;4^n]/6
当n=3k-1;4^n]/4 =R[n]-R[n]/: S[3k-1] =(-1/2[1^2+2^2+..: S[3k-2] =(-1/.;2 ∴S[3n]=n(9n+4)/.+(3k-1)^2]+3^3/4^2+1/4^2+2/:∵数列{a[n]}的通项a[n]=n^2[(cosnπ/2)(1/2)+2^2(-1/4^n]/.:S[3k]=n(3n+4)/.+[(9/4^3+;3-(2/12 =n(6n+8)/.;4^n]/2)[1^2+2^2+;4^2+.;3)^2].解.+(3k-2)^2]+3^3/2)[1^2+2^2+.+(3k)^2]+3^3/4^n)-n/2)(n/4^1)+2/4^3+;3)(1-1/2)R[n]+2(1/3) ∴S[n]=1^2(-1/2)+5^2(-1/.
提问者评价
其他类似问题
为您推荐:
数列的相关知识
等待您来回答
下载知道APP
随时随地咨询
出门在外也不愁}