设已知数列{an}满足a1=1,an-2an-1-2的n-1次方=0(n∈N*,n≥2).求证:数列{2的n次方分之an}是等差数列.若数列{an}的前n项和为sn,求sn.别用专业术语
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![设已知数列{an}满足a1=1,an-2an-1-2的n-1次方=0(n∈N*,n≥2).求证:数列{2的n次方分之an}是等差数列.若数列{an}的前n项和为sn,求sn.别用专业术语](/uploads/image/z/5466735-63-5.jpg?t=%E8%AE%BE%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%2Can%EF%BC%8D2an-1%EF%BC%8D2%E7%9A%84n-1%E6%AC%A1%E6%96%B9%EF%BC%9D0%EF%BC%88n%E2%88%88N%2A%2Cn%E2%89%A52%EF%BC%89.%E6%B1%82%E8%AF%81%EF%BC%9A%E6%95%B0%E5%88%97%7B2%E7%9A%84n%E6%AC%A1%E6%96%B9%E5%88%86%E4%B9%8Ban%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97.%E8%8B%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E6%B1%82sn.%E5%88%AB%E7%94%A8%E4%B8%93%E4%B8%9A%E6%9C%AF%E8%AF%AD)
设已知数列{an}满足a1=1,an-2an-1-2的n-1次方=0(n∈N*,n≥2).求证:数列{2的n次方分之an}是等差数列.若数列{an}的前n项和为sn,求sn.别用专业术语
设已知数列{an}满足a1=1,an-2an-1-2的n-1次方=0(n∈N*,n≥2).求证:数列{2的n次方分之an}是等差数列.若数列{an}的前n项和为sn,求sn.别用专业术语
设已知数列{an}满足a1=1,an-2an-1-2的n-1次方=0(n∈N*,n≥2).求证:数列{2的n次方分之an}是等差数列.若数列{an}的前n项和为sn,求sn.别用专业术语
a(n) = 2a(n-1) + 2^(n-1),
a(n+1) = 2a(n) + 2^n.
a(n+1)/2^(n+1) = a(n)/2^n + 1/2,
{a(n)/2^n}是首项为a(1)/2=1/2,公差为1/2的等差数列.
a(n)/2^n = 1/2 + (n-1)/2 = n/2.
a(n) = n*2^(n-1),
s(n) = a(1) + a(2) + a(3) + ... + a(n-1) + a(n)
= 1 + 2*2 + 3*2^2 + ... + (n-1)2^(n-2) + n2^(n-1).
2s(n) = 1*2 + 2*2^2 + ... + (n-1)2^(n-1) + n2^n.
s(n) = 2s(n) - s(n) = -1 - 2 - 2^2 - ... - 2^(n-1) + n2^n
= n2^n -[1+2+...+2^(n-1)]
= n2^n - [2^n - 1]/(2-1)
= n2^n - 2^n + 1
= 1 + (n-1)2^n