已知等差数列{an}满足a2=0,a6+a8=-10. (1)求数列{an}的通项公式 (2)求数列{an/2的n-1次方}的前n项和sn求过程,谢谢!
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![已知等差数列{an}满足a2=0,a6+a8=-10. (1)求数列{an}的通项公式 (2)求数列{an/2的n-1次方}的前n项和sn求过程,谢谢!](/uploads/image/z/1097304-24-4.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a2%3D0%2Ca6%2Ba8%3D-10.+%281%29%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F+%EF%BC%882%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Ban%2F2%E7%9A%84n-1%E6%AC%A1%E6%96%B9%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8Csn%E6%B1%82%E8%BF%87%E7%A8%8B%2C%E8%B0%A2%E8%B0%A2%21)
已知等差数列{an}满足a2=0,a6+a8=-10. (1)求数列{an}的通项公式 (2)求数列{an/2的n-1次方}的前n项和sn求过程,谢谢!
已知等差数列{an}满足a2=0,a6+a8=-10. (1)求数列{an}的通项公式 (2)求数列{an/2的n-1次方}的前n项和sn
求过程,谢谢!
已知等差数列{an}满足a2=0,a6+a8=-10. (1)求数列{an}的通项公式 (2)求数列{an/2的n-1次方}的前n项和sn求过程,谢谢!
1、∵{an}是等差数列
∴a6+a8=a2+4d+a2+6d
=2a2+10d
=-10
又∵a2=0
∴d=-1,a1=a2-d=1
则数列{an}的通项公式为:
an=a1+(n-1)d=2-n
2、an/2^(n-1)=(2-n)/2^(n-1)
则数列{an/2的n-1次方}的前n项和为:
sn=1/2^0+0-1/2^2-2/2^3-……-(2-n)/2^(n-1)
= 1-1/2^2-2/2^3-……-(1-n)/2^(n-2)-(2-n)/2^(n-1)
2sn=2-1/2^1-2/2^2-3/2^3-……-(2-n)/2^(n-2)
两式相减得:
sn=1-1/2^1-1/2^2-1/2^3-……-1/2^(n-2)-(2-n)/2^(n-1)
=1-1/2*[1-(1/2)^(n-2)]/(1-1/2)-(2-n)/2^(n-1)
=1-1+1/2^(n-2)-1/2^(n-2)+n/2^(n-1)
=n/2^(n-1)
a6+a8=2a7=-10
a7=-5
a2=0
d=-1
an=2-n
Tn=1/2^0+0/2^1+-1/2^2…………+(2-n)/2^(n-1) ①
1/2=1/2^1+0/2^2+-1/2^3…………+(2-n)/2^n ②
②-①得
-1/2Tn=-1/2^0+1/2^1+1/2^2+…………+1/2^(n-1)+(2-n)/2^n
Tn=n/2^(n-1)