P-BN20An evaluation of the accuracy of sonography for the measurements of gallbladder polyps across 3 UK hospital trusts: should the surveillance guidelines be revised?doi:10.1093/bjs/znab430.020BACKGROUND. Transabdominal ultrasound scan (USS) is recommended for surveillance of gallbladder polyps (...
已知等比数列{an}满足a1+a2=3,a2+a3=6.(1)求数列{an}的通项公式;(2)若bn=|an-20|,求数列{bn}的前n项和Tn.
已知{an}是等差数列.其中a10=30.a20=50.(1)求数列{an}的通项公式,(2)若bn=an-20.求数列{bn}的前n项和Tn的最小值.
可得b3-a3=20,a5+b2=20,设等差数列{an}的公差为d,正项等比数列{bn}的公比为q,q>0,则3q2-(3+2d)=3+4d+3q=20,解得d=2,q=3,所以(a_n)=2n+1,(b_n)=(3^n);(2)(c_n)=((2(((a_n)+3)))/(((a_n)(a_(n+1))(b_(n+1)))=(2(2n+4))/((2n+1)(2n+3)•3^(n...
【答案】(1)y由5=(a+1)2,且an0 242 n,当n=1时,,可求a1=1,当n=2时,S2=1+a2可求a2=3(2)由5=(a+1)2,且an0 242 n.可得当n≥2时,an=Sn-Sn-1=(+12(9n-1+1) 4 4可得an-an-1=2,结合等差数列的通项公式可求(3)由bn=20-an=21-2n可得Sn=-n2+20n=-(n-10)2+100,...
∴(an+an-1)(an-an-1-1)=0, ∴an-an-1=1, ∴数列{an}是以1为首项,公差为1的等差数列, ∴an=n; (2)由(1)知an=n, ∴bn=|n-20|, 当n≤20时,Tn=|1-20|+|2-20|+…+|n-20|=20n-(1+2+…+n)=39n−n2239n−n22; ...
解:(1)设等差数列{an}的公差为d,由题意可知\((array)la_1+d=3 5a_1+(5*4)/2*d=20(array).,解得\((array)la_1=2 d=1(array).,∴a_n=n+1(n∈N^*),当n≥2时,bn=Tn-Tn-1=1-bn-1+bn-1,∴(b_n)/(b_(n-1))=1/2,又T1=1-b1,∴b_1=1/2,∴{bn}是以1/2为首...
[答案] 36.查看解析[解析] 36.17.解:(1)设公差为d,公比为q,则 a2b2=(3+d)q=12 S3+b2=3a2+b2=3(3+d)+q=20 .2分 :.3d2-2d-21=0,(3d+7)(d-3)=0 ⊙{an}是单调递增的等差数列 ∴d0,∴d=3,g=2… 4分 ∴.an=3+(n-1)×3=3n,b=2 6分 Sn,n是偶数 ()Cn=S,cos...
解:(1)S_(20)=(a_1+a_2)+(a_3+a_4)+(a_5+a_6)+⋯+(a_(19)+a_(20))=1+5+9+⋯+37=((1+37)*10)/2=190.(2)a1=-2,an+an+1=2n-1 ①,an+2+an+1=2n+1②,②-①得an+2-an=2,∴bn+1-bn=a2n+2-a2n=2,∴b1=a2=3,数列{bn}是以3为首项,公差为2的等...
当n≥2时,由bn=2-2Sn,可得bn-bn-1=-2(Sn-Sn-1)=-2bn即 bn bn−1= 1 3…(4分)所以{bn}是以 b1= 2 3为首项, 1 3为公比的等比数列,于是 bn=2• 1 3n…(6分)(2)数列{an}为等差数列,公差 d= 1 2(a7−a5)=3,可得an=3n-1…(7分)从而 cn=an•bn=2(3n−1)• 1 3n...