a1=2,an+1=an+2^n,求数列An的通项公式和前n项和Sn【详细过程,
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![a1=2,an+1=an+2^n,求数列An的通项公式和前n项和Sn【详细过程,](/uploads/image/z/634627-19-7.jpg?t=a1%3D2%2Can%2B1%3Dan%2B2%5En%2C%E6%B1%82%E6%95%B0%E5%88%97An%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E5%92%8C%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E3%80%90%E8%AF%A6%E7%BB%86%E8%BF%87%E7%A8%8B%2C)
a1=2,an+1=an+2^n,求数列An的通项公式和前n项和Sn【详细过程,
a1=2,an+1=an+2^n,求数列An的通项公式和前n项和Sn【详细过程,
a1=2,an+1=an+2^n,求数列An的通项公式和前n项和Sn【详细过程,
a(n+1)=an+2^n
a(n+1)-an=2^n
an-a(n-1)=2^(n-1)
.
a3-a2=2^2
a2-a1=2^1
以上等式相加得
a(n+1)-a1=2^1+2^2+...+2^n
a(n+1)-a1=2*[1-2^n]/(1-2)
a(n+1)-a1=2^(n+1)-2
a(n+1)-2=2^(n+1)-2
a(n+1)=2^(n+1)
an=2^n
sn=a1+a2+.+an
=2^1+2^2+.+2^n
=2^1+2^2+.+2^n
=2*(1-2^n)/(1-2)
=2^(n+1)-2
【1】累加,an=2^n.n=1,2,....【2】Sn=2^(n+1)-2
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【1】累加,an=2^n.n=1,2,....【2】Sn=2^(n+1)-2.