已知数列{an}中,a1=5,an=2a(n-1)+2^n-1(n∈N*,n≥2) 若数列{an+λ/2^n}为等差数列
已知数列{an}中,a1=5,an=2a(n-1)+2^n-1(n∈N*,n≥2) 若数列{an+λ/2^n}为等差数列,求实数λ的值
正确答案是-1,可是我算到是1,
正确答案是-1,可是我算到是1,
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n≥2时,
an=2a(n-1)+2ⁿ -1
等式两边同除以2ⁿ
an/2ⁿ=a(n-1)/2^(n-1) +1 -1/2ⁿ
an/2ⁿ -1/2ⁿ=a(n-1)/2^(n-1) -1/2^(n-1) +1
(an -1)/2ⁿ -[a(n-1) -1]/2^(n-1)=1,为定值.
(a1-1)/2=(5-1)/2=2,数列{(an -1)/2ⁿ}是以2为首项,1为公差的等差数列.
(an+λ)/2ⁿ=(an -1)/2ⁿ=[an+(-1)]/2ⁿ
λ=-1,
an=2a(n-1)+2ⁿ -1
等式两边同除以2ⁿ
an/2ⁿ=a(n-1)/2^(n-1) +1 -1/2ⁿ
an/2ⁿ -1/2ⁿ=a(n-1)/2^(n-1) -1/2^(n-1) +1
(an -1)/2ⁿ -[a(n-1) -1]/2^(n-1)=1,为定值.
(a1-1)/2=(5-1)/2=2,数列{(an -1)/2ⁿ}是以2为首项,1为公差的等差数列.
(an+λ)/2ⁿ=(an -1)/2ⁿ=[an+(-1)]/2ⁿ
λ=-1,
1年前
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