设Sn是等差数列an的前n项和,a5=5a3,则S9/S5=?
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设Sn是等差数列an的前n项和,a5=5a3,则S9/S5=?
设Sn是等差数列an的前n项和,a5=5a3,则S9/S5=?
设Sn是等差数列an的前n项和,a5=5a3,则S9/S5=?
等差数列通项公式:an=a1+(n-1)d,前n项和公式:Sn=na1+n(n-1)d/2
所以:a5=a1+4d
a3=a1+2d
将a3、a5代人a5=5a3,解得:a1=-3d/2
所以:S9=9a1+36d=45d/2;S5=5a1+10d=5d/2
所以:S9/S5=9
(1)由于:{an}为等比数列则设公比为qi首项为a1由于前三项之积是64 (a2)^2=a1*a3则: (a2)^3=64 a2=4 又a2-18406a3-37395a4-9成等差数列 则:2(a3-3)=(a2-1)+(a4-9) 2(a2q-3)=3+(a2q^2-9) 2(4q-3)=3+(4q^2-9) q=0(舍)或q=2 则: an=a2...
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(1)由于:{an}为等比数列则设公比为qi首项为a1由于前三项之积是64 (a2)^2=a1*a3则: (a2)^3=64 a2=4 又a2-18406a3-37395a4-9成等差数列 则:2(a3-3)=(a2-1)+(a4-9) 2(a2q-3)=3+(a2q^2-9) 2(4q-3)=3+(4q^2-9) q=0(舍)或q=2 则: an=a2*q^(n-2) =4*2^(n-2) =2^n(2)bn=nan =n*2^n Sn=b1+b2+...+bn =1*2^1+2*2^2+......+n*2^n2Sn= 1*2^2+2*2^3+..+(n-1)*2^n+n*2^(n+1) -Sn=2^1+1*2^2+1*2^3+...+1*2^n-n*2^(n+1) =[2^1+2^2+...+2^n]-2n*2^n =[2(1-2^n)]/[1-2]-2n*2^n =2^(n+1)-2-2n*2^n =(2-2n)*2^n-2Sn=2-(2-2n)*2^n
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