由已知An=2A(n-1)+2^n-1(n属于正整数,n大于等于2)
得A4=2A3+15 ,可得A3=33
进而得A3=2A2+7,可得A2=13,
A2=2A+3 可得A1=5
当n>=2 时有
An=2A(n-1)+2^n-1
A(n-1)=2A(n-2)+2^(n-1)-1
A(n-2)=2A(n-3)+2^(n-2)-1
………………………
A2=2A1+2^2-1
将上面的n-1个等式变形为
An=2A(n-1)+2^n-1……………………(1)
2A(n-1)=2^2A(n-2)+2^n-2……………(2)
2^2A(n-2)=2^3A(n-3)+2^n-2^2……………(3)
……………………………………
2^(n-2)A2=2^(n-1)A1+2^n-2^(n-2)………(n-1)
对上面n-1个等式左右分别累加得
An=5*2^(n-1)+(n-1)2^n-[1+2+2^2+ …+2^(n-2)]
=(n+1)2^n+1
很显然,当n=1时A1=5也满足上式
故数列{An}得通项公式为An=(n+1)2^n+1
不难看出,取a= - 1
则(An+a)/2^n=n+1
数列{n+1}显然是等差数列
故存在a= -1使{(An+a)/2^n}为等差数列
∴Sn=[2+(n+1)]n/2
注:如要求严密,可再插入两部分内容,这里不再赘述:
(1)求出An=(n+1)2^n+1后,运用数学归纳法加以证明;
(2)得到(An+a)/2^n=n+1后用等差数列的定义证明数列{n+1}为等差数列.
已知数列{An}满足=2An-1+2^n-1(n属于正整数,n大于等于2)且A4=81.求数列{An
已知数列{An}满足=2An-1+2^n-1(n属于正整数,n大于等于2)且A4=81.求数列{An
已知数列{An}满足=2An-1+2^n-1(n属于正整数,n大于等于2)且A4=81.是否存在一个实数a,使数列{(an+a)/2^n}为等差数列?若存在求出a值,若不存在说明理由.
求前n项和Sn.
已知数列{An}满足=2An-1+2^n-1(n属于正整数,n大于等于2)且A4=81.是否存在一个实数a,使数列{(an+a)/2^n}为等差数列?若存在求出a值,若不存在说明理由.
求前n项和Sn.
数学人气:269 ℃时间:2020-03-28 08:46:04
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