等比数列{an}同时满足下列三个条件:a1+a6=33,a2a4=32,三个数4a2、2a3、a4依次成等差数列.
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等比数列{an}同时满足下列三个条件:a1+a6=33,a2a4=32,三个数4a2、2a3、a4依次成等差数列.
1)求数列{an}的通项公式
2)记bn=n/an求数列{bn}的前n项和Tn
是 a3a4=32
1)求数列{an}的通项公式
2)记bn=n/an求数列{bn}的前n项和Tn
是 a3a4=32
a(n)=aq^(n-1)
33=a(1)+a(6)=a+aq^5=a(1+q^5),
32=a(2)a(4)=aq*aq^3=a^2q^4,
4a(3)=4aq^2=4a(2)+a(4)=4aq+aq^3,0 = aq^3 - 4aq^2 + 4aq = aq(q^2 - 4q + 4 ) = aq(q-2)^2,q=2.
a=33/(1+q^5) = 33/(1+2^5)=1.
32=a^2q^4=16.矛盾.
不存在同时满足三个条件的等差数列a(n).
再问: Sorry!是a3a4=32
再答: 32=a(3)a(4)=aq^2aq^3=a^2q^5=1*2^5, a=1,q=2, a(n)=2^(n-1), b(n)=n/a(n)=n/2^(n-1). T(n)=1/2^(1-1) + 2/2^(2-1) + 3/2^(3-1) + ... + (n-1)/2^(n-2) + n/2^(n-1) 2T(n) = 1/2^(-1) + 2/2^(1-1) + 3/2^(2-1) +...+(n-1)/2^(n-3) + n/2^(n-2) T(n)=2T(n)-T(n)=1/2^(-1) + 1/2^(1-1) + 1/2^(2-1) + ... + 1/2^(n-2) - n/2^(n-1) = 2 + 1 + 1/2 + ... + 1/2^(n-2) - n/2^(n-1) = 2 - n/2^(n-1) + [1-1/2^(n-1)]/(1-1/2) = 2 - n/2^(n-1) + 2[1 - 1/2^(n-1)] = 4 - (n+2)/2^(n-1)
33=a(1)+a(6)=a+aq^5=a(1+q^5),
32=a(2)a(4)=aq*aq^3=a^2q^4,
4a(3)=4aq^2=4a(2)+a(4)=4aq+aq^3,0 = aq^3 - 4aq^2 + 4aq = aq(q^2 - 4q + 4 ) = aq(q-2)^2,q=2.
a=33/(1+q^5) = 33/(1+2^5)=1.
32=a^2q^4=16.矛盾.
不存在同时满足三个条件的等差数列a(n).
再问: Sorry!是a3a4=32
再答: 32=a(3)a(4)=aq^2aq^3=a^2q^5=1*2^5, a=1,q=2, a(n)=2^(n-1), b(n)=n/a(n)=n/2^(n-1). T(n)=1/2^(1-1) + 2/2^(2-1) + 3/2^(3-1) + ... + (n-1)/2^(n-2) + n/2^(n-1) 2T(n) = 1/2^(-1) + 2/2^(1-1) + 3/2^(2-1) +...+(n-1)/2^(n-3) + n/2^(n-2) T(n)=2T(n)-T(n)=1/2^(-1) + 1/2^(1-1) + 1/2^(2-1) + ... + 1/2^(n-2) - n/2^(n-1) = 2 + 1 + 1/2 + ... + 1/2^(n-2) - n/2^(n-1) = 2 - n/2^(n-1) + [1-1/2^(n-1)]/(1-1/2) = 2 - n/2^(n-1) + 2[1 - 1/2^(n-1)] = 4 - (n+2)/2^(n-1)
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