∴an=-46+(n-1)×4=4n-50,则bn=n(an+40)=n(4n-50+40)=4n2-10n.∵bn+1-bn=4(n+1)2-10(n+1)-4n2+10n=8n-6≥8×1-6=2>0.故数列{bn}是递增数列. 【分析】(1)设出等差数列的公差,由3a5=5a8得到公差和首项的关系,把前n项和用首项表示,配方后求得Sn取得最大值时n的值;(...
设等差数列{an},已知a5=-3,S10=-40(Ⅰ)求数列{an}的通项公式;(Ⅱ)若数列{abn}为等比数列,且b1=5,b2=8,求数列{bn}的前n项和Tn.
因为无穷数列{an}的各项均为互不相同的正整数,所以a1∈N*,d∈N*, 由a2=5,S5=40得,a1+d=5,5a1+5×425×42d=40, 解得a1=2,d=3,所以b2=S2a2S2a2-1=a1a2a1a2=2525; (2)因为数列{bn}为等差数列,所以2b2=b1+b3,即2(S2a2S2a2-1)=S1a1S1a1-1+S3a3S3a3-1, ...
分析:(Ⅰ)等差数列{an}中,由a5=-3,S10=-40,解得a1=5,d=-2.由此能求出数列{an}的通项公式.(Ⅱ)由{abn}为等比数列,b1=5,b2=8,知ab1=7-2b1=7-10=-3,ab2=7-2b2=7-16=-9,故abn=7-2bn=-3n,所以bn= 1 2•3n+ 7 2. cn= 37•an (2bn-7)2= 37•(7-2n) 9n=(7-2n)•...
而{bn}的首项b1=a1=1,公比满足q2= a3 a1= 9 1=9,得q=3∴bn=b1×3n-1=3n-1综上所述,数列{an}与{bn}的通项公式分别为an=4n-3、bn=3n-1;(2)由(1)得anbn=(4n-3)×3n-1∴Sn=1×1+5×31+9×32+…+(4n-7)×3n-2+(4n-3)×3n-1…①两边都乘以9,得3Sn=1×31+5×32+9×33...
解:(Ⅰ)设等差数列{an}的首项为a1、公差为d, ∵a5=-3,S10=-40, ∴解得:a1=5,d=-2. ∴an=7-2n. (Ⅱ)由(Ⅰ)知,an=7-2n,又数列为等比数列,且b1=5,b2=8, ∴q===3, 又AD_1=a5=7-2×5=-3, ∴AD_n=(-3)×3n-1=-3n,又AD_n=7-2bn, ∴7-2bn=-3n, ∴bn=+, ∴数列{b...
解:(1)设公差为d,则由a2=5,S5=40,得:{a1+d=5a1+2d=8,解得{a1=2d=3,则an=3n-1…(2)∵q3=b4b1=813=27∴q=3bn=b1qn−1=3∙3n−1=3n…(3)Tn=c1+c2+c3+…+cn=2×3+5×32+8×33+…+(3n−1)3n①①-②:−2Tn=2×3+3(32+33+…+3n)−(3n−1)3n+1...
设a1=x;a2=x+y=5;s5=5x+10y=40;x=2;y=3;an=2+3(n-1);an=3n-1;设bn=3*q^(n-1);b4=3*q^3=81;q=3;bn=3ⁿ解
an=Sn-S(n-1)=n²+n-(n-1)²-(n-1)=2nn=1时也满足∴ an=2n二、bn=(n+1)/[(n+2)²*(2n)²]=[4(n+1)/16]/[(n+2)²*n²]=(1/16)*[1/n² -1/(n+2)²]∴ Tn=(1/16)*[1/1-1/9+1/4-1/16+.+1/(n-1)²-1/(n+1)²+1/n² -1/(n+2)...
4.已知两个等差数列{an}.{bn}.它们的前n项和分别是Sn.Tn.若$\frac{S n}{T n}$=$\frac{2n+3}{3n-1}$.则$\frac{{{a 3}+{a 5}}}{{{b 3}+{b 5}}}$+$\frac{a 4}{{{b 2}+{b 6}}}$=$\frac{51}{40}$.