高一数学，an1通项公式怎么求？ an=2*an-1+1的通项公式怎么求
\u9ad8\u4e00\u6570\u5b66\uff0ca1=2,an+1=an+2^n\uff0c\u6c42\u6570\u5217An\u7684\u901a\u9879\u516c\u5f0f\u548c\u524dn\u9879\u548cSn\u3010\u8be6\u7ec6\u8fc7\u7a0b\uff0c\u8c22\u8c22\u3011a(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
\u4ee5\u4e0a\u7b49\u5f0f\u76f8\u52a0\u5f97
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
\u65b9\u6cd5\u4e00
an + 1 = 2[a(n-1) + 1]
an + 1 = 2^(n-1) * (a1 + 1)
an = 2^(n-1) * (a1 + 1) - 1
\u65b9\u6cd5\u4e8c
an/2^n = a(n-1)/2^(n-1) + 1/2^n
an/2^n = a1/2 + 1/2 - 1/2^n
an = 2^(n-1) * a(n-1) + 2^(n-1) - 1
\u6269\u5c55\u8d44\u6599\u6c42\u901a\u9879\u65b9\u6cd5
\uff081\uff09\u5f85\u5b9a\u7cfb\u6570\u6cd5\uff1a\u5df2\u77e5a\uff08n+1\uff09=2an+3\uff0ca1=1\uff0c\u6c42an
\u6784\u9020\u7b49\u6bd4\u6570\u5217a\uff08n+1\uff09+x=2\uff08an+x\uff09
a\uff08n+1\uff09=2an+x\uff0c\u2235a\uff08n+1\uff09=2an+3 \u2234x=3
\u2234\uff08a\uff08n+1\uff09+3\uff09/\uff08an+3\uff09=2
\u2234{an+3}\u4e3a\u9996\u9879\u4e3a4\uff0c\u516c\u6bd4\u4e3a2\u7684\u7b49\u6bd4\u6570\u5217\uff0c\u6240\u4ee5an+3=a1*q^(n-1)=4*2^(n-1),an=2^(n+1)-3
\uff082\uff09\u5b9a\u4e49\u6cd5\uff1a\u5df2\u77e5Sn=a\u00b72^n+b,\uff0c\u6c42an\u7684\u901a\u9879\u516c\u5f0f\uff1f
\u2235Sn=a\u00b72^n+b\u2234Sn-1=a\u00b72^n-1+b
\u2234an=Sn-Sn-1=a\u00b72^n-1 \u3002
a1=1
for n>=2
3Sn -4, an , 2- (3/2)S(n-1) 成等差数列
To find : an
Solution
3Sn -4, an , 2- (3/2)S(n-1) 成等差数列
2an = 3Sn -4 + 2- (3/2)S(n-1)
2[Sn -S(n-1)] = 3Sn - (3/2)S(n-1) -2
Sn = -(1/2)S(n-1) +2
Sn - 4/3 = -(1/2) [ S(n-1) -4/3 ]
=> {Sn - 4/3} 是等比数列 , q= -1/2
Sn - 4/3 =(-1/2)^(n-1) . (S1 - 4/3)
Sn = 4/3 +(-1/3). (-1/2)^(n-1)
an = Sn -S(n-1)
=(-1/3). (-1/2)^(n-1) - (-1/3). (-1/2)^(n-2)
=(-1/3) (-1/2)^(n-2). ( -1/2 - 1)
= -(-1/2)^(n-1)
a1=1 满足 （1）
ie
an= -(-1/2)^(n-1)
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