已知正数数列an中,an^2-ana(n-1)^2-2a(n-1)^2=0(n≥2),a1=1(1)求数列an的通项公式(2)设bn=nan,求数列bn的前n项和Tn注:(n-1)是指n-1项
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![已知正数数列an中,an^2-ana(n-1)^2-2a(n-1)^2=0(n≥2),a1=1(1)求数列an的通项公式(2)设bn=nan,求数列bn的前n项和Tn注:(n-1)是指n-1项](/uploads/image/z/9317835-27-5.jpg?t=%E5%B7%B2%E7%9F%A5%E6%AD%A3%E6%95%B0%E6%95%B0%E5%88%97an%E4%B8%AD%2Can%5E2-ana%28n-1%29%5E2-2a%28n-1%29%5E2%3D0%28n%E2%89%A52%EF%BC%89%2Ca1%3D1%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%882%EF%BC%89%E8%AE%BEbn%3Dnan%2C%E6%B1%82%E6%95%B0%E5%88%97bn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn%E6%B3%A8%EF%BC%9A%EF%BC%88n-1%29%E6%98%AF%E6%8C%87n-1%E9%A1%B9)
已知正数数列an中,an^2-ana(n-1)^2-2a(n-1)^2=0(n≥2),a1=1(1)求数列an的通项公式(2)设bn=nan,求数列bn的前n项和Tn注:(n-1)是指n-1项
已知正数数列an中,an^2-ana(n-1)^2-2a(n-1)^2=0(n≥2),a1=1
(1)求数列an的通项公式
(2)设bn=nan,求数列bn的前n项和Tn
注:(n-1)是指n-1项
已知正数数列an中,an^2-ana(n-1)^2-2a(n-1)^2=0(n≥2),a1=1(1)求数列an的通项公式(2)设bn=nan,求数列bn的前n项和Tn注:(n-1)是指n-1项
1.a²n-an*na(n-1)-2a²(n-1)=0
(an-2a(n-1))(an+a(n-1))=0
正数数列an中,所以an>0,an+a(n+1)>0
an-2a(n-1)=0
an=2a(n-1)
an=a1*2^(n-1)
an=2^(n-1)
2.bn=nan,bn=n*2^(n-1)
tn=1*2^0+2*2^1+...+n2^(n-1)
2tn=1*2^1+...+(n-1)*2^(n-1)+n*2^n
2tn-tn=-1*2^0-2^1-2^2-...-2^(n-1)+n*2^n
tn=-(2^0+2^1+...+2^(n-1))+n*2^n
tn=-2^n+1+n*2^n
tn=(n-1)*2^n+1
递推式“an^2-ana(n-1)^2-2a(n-1)^2=0”中的第二项“ana(n-1)^2”最后那个平方究竟有没有?一楼是按没有做的,是题目写错了,还是一楼看错了?