已知数列an满足a1=1/2,2an+1-an=1.1.求an的通项公式 2.证明:求和Sn=a已知数列an满足a1=1/2,2an+1-an=1.1.求an的通项公式2.证明:求和Sn=a1+a2+···+an
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![已知数列an满足a1=1/2,2an+1-an=1.1.求an的通项公式 2.证明:求和Sn=a已知数列an满足a1=1/2,2an+1-an=1.1.求an的通项公式2.证明:求和Sn=a1+a2+···+an](/uploads/image/z/1549804-4-4.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D1%2F2%2C2an%2B1-an%3D1.1.%E6%B1%82an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F+2.%E8%AF%81%E6%98%8E%3A%E6%B1%82%E5%92%8CSn%3Da%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D1%2F2%2C2an%2B1-an%3D1.1.%E6%B1%82an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F2.%E8%AF%81%E6%98%8E%3A%E6%B1%82%E5%92%8CSn%3Da1%2Ba2%2B%C2%B7%C2%B7%C2%B7%2Ban)
已知数列an满足a1=1/2,2an+1-an=1.1.求an的通项公式 2.证明:求和Sn=a已知数列an满足a1=1/2,2an+1-an=1.1.求an的通项公式2.证明:求和Sn=a1+a2+···+an
已知数列an满足a1=1/2,2an+1-an=1.1.求an的通项公式 2.证明:求和Sn=a
已知数列an满足a1=1/2,2an+1-an=1.
1.求an的通项公式
2.证明:求和Sn=a1+a2+···+an
已知数列an满足a1=1/2,2an+1-an=1.1.求an的通项公式 2.证明:求和Sn=a已知数列an满足a1=1/2,2an+1-an=1.1.求an的通项公式2.证明:求和Sn=a1+a2+···+an
2a(n+1) = a(n) + 1,
2a(n+1) - 2 = a(n) - 1,
a(n+1)-1 = [a(n)-1]/2,
{a(n)-1}是首项为a(1)-1=-1/2,公比为1/2的等比数列.
a(n)-1 = (-1/2)(1/2)^(n-1) = -1/2^n,
a(n) = 1 -1/2^n
s(n) = n - [1/2 + 1/2^2 + ...+ 1/2^n]
= n - (1/2)[1 + 1/2 + ...+ 1/2^(n-1)]
= n - (1/2)[1 - 1/2^n]/(1-1/2)
= n - 1 + 1/2^n
题出错了,2an+1-an=1.这玩意。。。。不就是说an=0么。。。
2a(n+1) = a(n) + 1,
2a(n+1) - 2 = a(n) - 1,
a(n+1)-1 = [a(n)-1]/2,
{a(n)-1}是首项为a(1)-1=-1/2,公比为1/2的等比数列。
a(n)-1 = (-1/2)(1/2)^(n-1) = -1/2^n,
a(n) = 1 -1/2^n
s(n) = n - [1/2 + 1...
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2a(n+1) = a(n) + 1,
2a(n+1) - 2 = a(n) - 1,
a(n+1)-1 = [a(n)-1]/2,
{a(n)-1}是首项为a(1)-1=-1/2,公比为1/2的等比数列。
a(n)-1 = (-1/2)(1/2)^(n-1) = -1/2^n,
a(n) = 1 -1/2^n
s(n) = n - [1/2 + 1/2^2 + ... + 1/2^n]
= n - (1/2)[1 + 1/2 + ... + 1/2^(n-1)]
= n - (1/2)[1 - 1/2^n]/(1-1/2)
= n - 1 + 1/2^n
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