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

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已知等差数列{an}满足a2=0a6+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项和已知等差数列{an}满足a2=0a6+a8=-1

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
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

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
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﹚]

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) ...

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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)]

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