设数列{an}为等比数列,数列{bn}满足bN=na1+(n-1)a2+…+2an-1+an ps:只需第三问!须详述!
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设数列{an}为等比数列,数列{bn}满足bN=na1+(n-1)a2+…+2an-1+an ps:只需第三问!须详述!
设数列{an}为等比数列,数列{bn}满足bN=na1+(n-1)a2+…+2an-1+an,n属于正整数.已知b1=m,b2=3m/2,其中m不等于0
(1)求数列{an}的首项和公比;
(2)当m=1时,求bn;
(3)设Sn为数列{an}的前n项和,若对于任意的正整数n,都有Sn属于[1,3],求实数m的取值范围
设数列{an}为等比数列,数列{bn}满足bN=na1+(n-1)a2+…+2an-1+an,n属于正整数.已知b1=m,b2=3m/2,其中m不等于0
(1)求数列{an}的首项和公比;
(2)当m=1时,求bn;
(3)设Sn为数列{an}的前n项和,若对于任意的正整数n,都有Sn属于[1,3],求实数m的取值范围
![设数列{an}为等比数列,数列{bn}满足bN=na1+(n-1)a2+…+2an-1+an ps:只需第三问!须详述!](/uploads/image/z/6818942-38-2.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E6%95%B0%E5%88%97%7Bbn%7D%E6%BB%A1%E8%B6%B3bN%3Dna1%2B%28n-1%29a2%2B%E2%80%A6%2B2an-1%2Ban+ps%3A%E5%8F%AA%E9%9C%80%E7%AC%AC%E4%B8%89%E9%97%AE%21%E9%A1%BB%E8%AF%A6%E8%BF%B0%21)
a(n)=aq^(n-1),
b(n) = na(1)+(n-1)a(2)+...+2a(n-1)+a(n),
m=b(1)=a(1)=a,
3m/2 = b(2) = 2a(1)+a(2) = 2m + a(2), a(2)=-m/2.
q = a(2)/a(1)=-1/2
a(n) = m*(-1/2)^(n-1).
m=1时,a(n) = (-1/2)^(n-1).
b(n) = n*1 + (n-1)(-1/2) + (n-2)(-1/2)^2 + ... + 2(-1/2)^(n-2) + (-1/2)^(n-1),
-2b(n) = n*(-2) + (n-1)*1 + (n-2)(-1/2) + ... + 2(-1/2)^(n-3) + (-1/2)^(n-2),
3b(n) = b(n) - [-2b(n)] = 2n + 1 + (-1/2) + ... + (-1/2)^(n-2) + (-1/2)^(n-1)
= 2n + [1-(-1/2)^n]/[1-(-1/2)]
= 2n + (2/3)[1 - (-1/2)^n],
b(n) = 2n/3 + 2[1 - (-1/2)^n]/9
a(n) = m(-1/2)^(n-1),m不为0.
s(n) = a(1)+a(2)+...+a(n) = m[1+(-1/2)+...+(-1/2)^(n-1)] = m[1 - (-1/2)^n]/[1-(-1/2)]
= (2m/3)[ 1 - (-1/2)^n],
s(2n-1) = (2m/3) - (2m/3)(-1/2)^(2n-1) = 2m/3 + (2m/3)(1/2)^(2n-1), 单调递减. 2m/3 s(2n) >= m/2.
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b(n) = na(1)+(n-1)a(2)+...+2a(n-1)+a(n),
m=b(1)=a(1)=a,
3m/2 = b(2) = 2a(1)+a(2) = 2m + a(2), a(2)=-m/2.
q = a(2)/a(1)=-1/2
a(n) = m*(-1/2)^(n-1).
m=1时,a(n) = (-1/2)^(n-1).
b(n) = n*1 + (n-1)(-1/2) + (n-2)(-1/2)^2 + ... + 2(-1/2)^(n-2) + (-1/2)^(n-1),
-2b(n) = n*(-2) + (n-1)*1 + (n-2)(-1/2) + ... + 2(-1/2)^(n-3) + (-1/2)^(n-2),
3b(n) = b(n) - [-2b(n)] = 2n + 1 + (-1/2) + ... + (-1/2)^(n-2) + (-1/2)^(n-1)
= 2n + [1-(-1/2)^n]/[1-(-1/2)]
= 2n + (2/3)[1 - (-1/2)^n],
b(n) = 2n/3 + 2[1 - (-1/2)^n]/9
a(n) = m(-1/2)^(n-1),m不为0.
s(n) = a(1)+a(2)+...+a(n) = m[1+(-1/2)+...+(-1/2)^(n-1)] = m[1 - (-1/2)^n]/[1-(-1/2)]
= (2m/3)[ 1 - (-1/2)^n],
s(2n-1) = (2m/3) - (2m/3)(-1/2)^(2n-1) = 2m/3 + (2m/3)(1/2)^(2n-1), 单调递减. 2m/3 s(2n) >= m/2.
1
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