Question

已知等差数列{an}满足a2=0 a6+a8=-10 求数列{an}的通项公式 数列{an/2的n-1次方}的前n项和

求数列{an/2的n-1次方}的前n项和

Answer 1

a6+a8=2a7=-10
∴a7=-5
a2=0
∴d=﹙﹣5－0﹚÷﹙7-5﹚=-1
∴an=﹣n+2
数列{an/2的n-1次方}的前n项和(错位相减法）
s=1＋0-1／2²-2／2³＋……－﹙n－2﹚／[2^﹙n-1﹚]
½s=½＋0－1／2³＋……－﹙n－1﹚／[2^﹙n-1﹚]－﹙n－2﹚／[2^﹙n﹚]
s-½s=½-1／2²－1／2³＋……－1／[2^﹙n-1﹚]＋﹙n－2﹚／[2^﹙n﹚]
½s=½-﹛1／2²＋1／2³＋……＋1／[2^﹙n-1﹚]﹜＋﹙n－2﹚／[2^﹙n﹚]
=½－﹛1／2²－1／[2^﹙n﹚]﹜／﹙1-½﹚＋﹙n－2﹚／[2^﹙n﹚]
=½－1／2＋2／[2^﹙n﹚]＋﹙n－2﹚／[2^﹙n﹚]
=n／[2^﹙n﹚]
∴s=n／[2^﹙n-1﹚]

Answer 2

a6+a8=a7+a7=-10
a7=-5
a2=0
d=-1
an=-n+2
后面一题题目没看懂

追问

求数列{an/2的n-1次方}的前n项和

追答

不会解

Answer 3

a6 +a8 = 2a7 = -10
a7 = -5 5d = a7 -a2 =-5 -0 =-5
公差d=-1
an= a2 + (n-2)d = 2-n

an/2的n-1次方:an/[2^(n-1)]
Tn = a1/(2^0) + a2/(2^1) + a3/(2^2) +a4/(2^3) +···+an/[2^(n-1)] ········································（1）
Tn/2= a1/(2^1)+ a2/(2^2) +a3/(2^) + ···+a(n-1)/[2^(n-1)] + an/[2^n]···························（2）
（1）-（2）: Tn/2 = a1 + (a2-a1)/(2^1) +(a3-a2)/(2^2) +(a4-a3)/(2^3) +···+ [an-a(n-1)] /[2^(n-1)]
-an/(2^n) = a1-an/(2^n) +d[(1/2)^1+(1/2)^2+(1/2)^3+(1/2)^4 + ···+(1/ 2)^(n-1)]
= 1 - (2-n)/(2^n) - (1/2)[1-(1/2)^(n-1)]/(1-1/2)
= 1-(2-n)/(2^n)- [1-(1/2)^(n-1)]
= n/(2^n)
所以Tn = n/[2^(n-1)]

Answer 4

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