已知等差数列an首项为1,公差不等于0,等比数列bn前三项满足b1=a1*b2...答:应该是b1=a1,b2=a2,b3=a6吧设公差为da1 = 1a2 = 1+da6 = 1+5d则(1+5d)=(1+d)^2d不等于0,d=3an = 3n-2Sn = (3n - 1)n/2m(an+1)- Sn = 4,等号当3n-1 = 12时取得,但n是整数所以最小值在n=4或5的时候取得,经计算,最小值为n=4时取得,所以24/(3n-1)+ n/2 >=...
已知数列{An}满足A1=1,An+1={1/2An+n-1,n为奇数,An-2n,n为偶数},记B...答:已知数列{An}满足A1=1,An+1={1/2An+n-1,n为奇数,An-2n,n为偶数},记Bn=A2nn∈N* 求数列{Bn}的通项公式(提示:Bn+1=A2(n+1)=A(2n+1)+1)设Cn=(2^2n-1-1)Bn²,数列{Cn}的前n项和为Sn,若对任意n属于N*,不等式入大于等于1+Sn恒成立,求实数入的取值范围。... 求数列{Bn}的通项...
a1等于5,公差规律是3 5 7 9的数列怎求an啊 也就是说a2等于8,a3等于13...答:解:数列an 相邻项差值满足等差数列{bn},b1=3,d=2.则bn=b1+(n-1)d=2n+1,有: a2-a1=b1,……a<n>-a<n-1>=b<n-1>,即:a2-a1 = 2* 1+1,a3-a2 = 2* 2+1,……a<n-1>-a<n-2>=2* (n-2) +1,a<n>- a<n-1> = 2* (n-1) +1,上列式子左右分别相加有:a...