数列an的通项an=n&sup2;(cos&sup2;nπ/3-sin&sup2;nπ/3) 前n项和为Sn(1)求Sn（2）bn=S3n/n×4^n(分母是n乘4的n次方）求数列bn的前n项和Tn

首页 > 中学数学试题 > 题目详情

数列an的通项an=n²(cos²nπ/3-sin²nπ/3) 前n项和为Sn(1)求Sn（2）bn=S3n/n×4^n(分母是n乘4的n次方）求数列bn的前n项和Tn

2022-10-07 17:00:02 14次反馈错误加入收藏正确率 : 100%

题目内容：

数列an的通项an=n²(cos²nπ/3-sin²nπ/3) 前n项和为Sn
(1)求Sn（2）bn=S3n/n×4^n(分母是n乘4的n次方）求数列bn的前n项和Tn

优质答案解析

王老师

回答题目：2621条

∵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
(2)∵当n=3k时：S[3k]=n(3n+4)/6=3k(9k+4)/6=k(9k+4)/2
∴S[3n]=n(9n+4)/2
∵b[n]=S[3n]/(n4^n)
∴b[n]=(9n+4)/(2*4^n)=(9/2)(n/4^n)+2/4^n
设R[n]=1/4^1+2/4^2+3/4^3+...+n/4^n
则R[n]/4=1/4^2+2/4^3+3/4^4+...+n/4^(n+1)
∴3R[n]/4
=R[n]-R[n]/4
=(1/4^1+1/4^2+1/4^3+...+1/4^n)-n/4^(n+1)
=(1/4)(1-1/4^n)/(1-1/4)-n/4^(n+1)
=(1/3)(1-1/4^n)-n/4^(n+1)
=[4-(3n+4)/4^n]/12
∴R[n]=[4-(3n+4)/4^n]/9
∴T[n]
=[(9/2)(1/4^1)+2/4^1]+[(9/2)(2/4^2)+2/4^2]+...+[(9/2)(n/4^n)+2/4^n]
=(9/2)(1/4^1+2/4^2+...+n/4^n)+2(1/4^1+1/4^2+1/4^3+...+1/4^n)
=(9/2)R[n]+2(1/4)(1-1/4^n)/(1-1/4)
=[4-(3n+4)/4^n]/2+(2/3)(1-1/4^n)
=2-(3n/2+2)/4^n+2/3-(2/3)/4^n
=[16-(9n+16)/4^n]/6

本题链接：

大家查题吧

首页 > 中学数学试题 > 题目详情

数列an的通项an=n²(cos²nπ/3-sin²nπ/3) 前n项和为Sn(1)求Sn（2）bn=S3n/n×4^n(分母是n乘4的n次方）求数列bn的前n项和Tn

题目内容：

数列an的通项an=n²(cos²nπ/3-sin²nπ/3) 前n项和为Sn
(1)求Sn（2）bn=S3n/n×4^n(分母是n乘4的n次方）求数列bn的前n项和Tn

看完仍有疑问？有类似问题直接问老师

大家查题吧

首页 > 中学数学试题 > 题目详情