设数列{an}是首项为50,公差为2的等差数列;{bn}是首项为10,公差为4的等差数列,以ak、bk为相邻两边的矩形内最大圆面积记为Sk,则Sk等于. 相关知识点: 代数 数列 数列与解析几何的综合 试题来源: 解析(2k+3)2π(k≤21) (k+24)2π(k>21)【分析】根据数列{an}是首项为50,公差为2的等差数列...
又n=1时,a1=S1=10-1=9,满足上式; ∴an=11-2n, ∴bn=|an|=|11-2n|; 显然n≤5时,bn=an=11-2n,Tn=10n-n2; n≥6时,bn=-an=2n-11, ∴Tn=(a1+a2+…+a5)-(a6+a7+…+an)=2S5-Sn=50-10n+n2 故Tn={10n−n2,n≤5n2−10n+50,n≥6{10n−n2,n≤5n2−10n+50,n≥6 ...
由a10=30,a20=50,得方程组{a1+9d=30,a1+19d=50, 解得a1=12,d=2. 所以an=12+2(n-1)=2n+10. (2)证明:由(1)得,bn=2an−10=22n+10-10=22n=4n, 所以bn+1bn=4n+14n=4. 所以{bn}为首项为4,公比为4的等比数列. (3)由nbn=n×4n,得 Tn=1•4+2•42+…+n•4n,① 4Tn...
bn-1 =q(q为不等0的常数) 解答:解:(1)由an=a1+(n-1)d,a10=30,a20=50,得方程组 a1+9d=30 a1+19d=50 ,…(2分) 解得a1=12,d=2.…(4分) ∴an=12+(n-1)•2=2n+10.…(5分) (2)由Sn=na1+ n(n-1) 2 d,Sn=210…(7分) ...
解：设公差为d。(1)a20-a10=10d=50-30=20 d=2 a1=a10-9d=30-2×9=12 an=a1+(n-1)d=12+2(n-1)=2n+10 数列{an}的通项公式为an=2n+10。(2)题出错了，是等差数列吧。bn=2an-10=2×(2n+10)-10=4n-10 b1=4×1-10=-6 b(n+1)-bn=4(n+1)-10-4n+10=4，为定值...
解:(1)设数列{an}的公差为d,则an=a1+(n-1)d, 由a10=30,a20=50,得方程组{a1+9d=30,a1+19d=50, 解得a1=12,d=2. 所以an=12+2(n-1)=2n+10. (2)证明:由(1)得,bn=2an−10=22n+10-10=22n=4n, 所以bn+1bn=4n+14n=4. 所以{bn}为首项为4,公比为4的等比数列. (3)由nbn...
a10=30,a20=50.a20=a10+(20-10)d 50=30+10d d=2 所以 an=a10+(n-10)d =30+2(n-10)即 an=2n+10 (2)bn=2^(an-10)=2^(2n)=4^n (应该是这样的吗)bn/bn-1=4 所以 {bn}是以4为公比的等比数列。a10
a10=a1+9d a20=a1+19d 想减,10d=20 d=2,a1=12 an=a1+(n-1)d=2n+10 bn=2^(an-10)=2^(2n)=4^n bn为等比数列 前n项和为a1(1-q^n)/(1-q)=4(4^n-1)/3 第三题 裂项求和 4/(an-10)(an-8)=4/(2n)(2n+2)=1/n(n+1)=1/n-1/(n+1) 前n项和为1/1-1/(n+1)=...
IBM confirms liking for yen with 50bn yen 10 year outing.Reports that computer company IBM has tapped the yen denominated market for a fourth time with a bond issue in 2000. Price of the deal; Details on the deal's allocation; Criticism to the deal.EBSCO_bspEuroweek...
【答案】分析:(1)利用等差数列的通项公式分别表示出前四项和与a2,a3,a7等比数列关系组成方程组求得a1和d,最后根据等差数列的通项公式求得an. (2)把(1)中求得的an代入 中,可知数列{bn}为等比数列,进而根据等比数列的求和公式求得答案. 解答:解:(1)由题意知 ...