数列{an}满足a(n+2)-2a(n+1)+an=4,a1=1,a2=3.求其通项公式
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/06 10:14:20
xn@_ŊT]\9v81Ѝ@HUQ*H\R@jH/MS_]zȍ9;_Qz~~UZ/xBS[dG/&&l&fl#-P|,Mliqj^|AF5Ͷ
数列{an}满足a(n+2)-2a(n+1)+an=4,a1=1,a2=3.求其通项公式
数列{an}满足a(n+2)-2a(n+1)+an=4,a1=1,a2=3.求其通项公式
数列{an}满足a(n+2)-2a(n+1)+an=4,a1=1,a2=3.求其通项公式
你好 具体过程如图:
a(n+2)-2a(n+1)+an=4
a(n+2)-a(n+1) = a(n+1) - a(n) +4
令b(n) = a(n+1)-a(n)
得b(n+1) = b(n) + 4
b(n) = 4*(n-1) + b(1) = 4n-4+2 = 4n-2
因此:
a(n+1) - a(n) = 4n - 2
a(n) - a(n-1)...
全部展开
a(n+2)-2a(n+1)+an=4
a(n+2)-a(n+1) = a(n+1) - a(n) +4
令b(n) = a(n+1)-a(n)
得b(n+1) = b(n) + 4
b(n) = 4*(n-1) + b(1) = 4n-4+2 = 4n-2
因此:
a(n+1) - a(n) = 4n - 2
a(n) - a(n-1) = 4(n-1) - 2
....
a(2) - a(1) = 4*1 - 2
上述n式相加,得 a(n+1) - a(1) = 4*(1+...+n) - 2n = 4*n(n+1)/2 - 2n = 2n^2
因此 a(n) = 2*(n-1)^2 + a(1) = 2n^2 - 4n + 3
收起
数列{an}满足a1=2,a(n+1)=2an+n+2,求an
通项公式为an=a(n^2)+n的数列{an},若满足a1
周期性数列问题i已知数列{an}满足a(n+1)=2an (0
已知数列an满足a1=2,an=a(n-1)+2n,(n≥2),求an
已知数列{a}满足a1=1/2,a(n+1)=an+1/(n^2+n),求an已知数列{a}满足a1=1/2,a(n+1)=an+1/(n^2+n),求an
已知数列{an}满足a1=33,a(n+1)-an=2n,则an/n的最小值
已知数列{an}满足a1=33,a(n+1)-an=2n,求an/n的最小值
已知数列an满足:a1=1,an-a(n-1)=n n大于等于2 求an
已知数列an满足a1=100,a(n+1)-an=2n,则(an)/n的最小值为
难题!数列an满足啊a(n+3)》n+3,a(n+2)》n+2,求a2007的值
数列[An]满足a1=2,a(n+1)=3an-2 求an
数列an满足a1=1,a(n+1)=an/[(2an)+1],求a2010
数列{An}满足A1=1,A(n+3)=An+3,A(n+2)=An +2
已知数列{an}满足a1=2,a(n+1)-an=a(n+1)*an,则a31=?
求一道很简单的数列题数列{a}满足an=(n^2+n+1)/3求an+1
已知数列{an}满足条件a1=3,且a( n+1)-an=(20)^n+n,求通项公式已知数列{an}满足条件a1=3,且a( n+1)-an=(2)^n+n,求通项公式
已知数列{an}满足a1=a,a2=b,a(n+1)=a(n+2)+an,求a2012
已知数列an中,满足a1=6a,a(n+1)+1=2[(an)+1],n属于N*,求数列an的通项公式