作业帮已知等差数列{an}满足a2=-7,a6+a8=6,求数列{ an/2^n-1 }的前n项和?
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已知等差数列{an}满足a2=-7,a6+a8=6,求数列{ an/2^n-1 }的前n项和?
已知等差数列{an}满足a2=-7,a6+a8=6,求数列{ an/2^n-1 }的前n项和?
已知等差数列{an}满足a2=-7,a6+a8=6,求数列{ an/2^n-1 }的前n项和?
已知等差数列{an}满足a2=-7,a6+a8=6,有
a2+4d+a2+6d=2a2+10d=6
得d=[6-2*(-7)]/10=2
得an=a2+(n-2)d=-7+2(n-2)=2n-11
an/2^( n-1)=(2n-11)/2^( n-1)=n/2^(n-2)-11/2^( n-1)
Sn=1/(1/2)+2/1+3/2+...+n/2^(n-2)
2Sn=2/(1/2)+3/1+...+n/2^(n-3)
得Sn=1/(1/2)+1/1+...+1/2^(n-3)-n/2^(n-2)=4-1/2^(n-3)-n/2^(n-2)
数列{an/2n-1}的前n项和
=4-1/2^(n-3)-n/2^(n-2)-11[2-1/2^(n-2)]
=-18-1/2^(n-3)-n/2^(n-2)+11/2^(n-2)
=-18+/2^(n-2)-n/2^(n-2)
a6=a2+4d
a8=a2+6d
a6+a8=2*a2+10d=6
d=2
a1=-9
an=a1+(n-1)d=2n-10
an/2^n-1=(2n-10)/2^n-1=(n-5)/2^n-2
设s=an/2^n-1 则
s=-4/2^-1+(-3)/2^0+(-2)/2^1+....+(n-5)/2^n-2
s/2= ...
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a6=a2+4d
a8=a2+6d
a6+a8=2*a2+10d=6
d=2
a1=-9
an=a1+(n-1)d=2n-10
an/2^n-1=(2n-10)/2^n-1=(n-5)/2^n-2
设s=an/2^n-1 则
s=-4/2^-1+(-3)/2^0+(-2)/2^1+....+(n-5)/2^n-2
s/2= -4/2^0+(-3)/2^1+(-2)/2^2+....+(n-5)/2^n-1
s-s/2=-8+1/2^0+1/2^1+1/2^2+.....+1/2^n-2-(n-5)/2^n-1
s/2=-8+1+(1/2)^1+(1/2)^2+....++(1/2)^n-2-(n-5)/2^n-1
s/2=-8+(1*(1-(1/2)^n-1))/(1-1/2)-(n-5)/2^n-1
s=-12-(1/2)^n-3-(n-5)/2^n-2
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