已知等差数列an与bn的前n项和分别为Sn和Tn (3n-2)╱(2n 1)
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![已知等差数列an与bn的前n项和分别为Sn和Tn (3n-2)╱(2n 1)](/uploads/image/f/4277973-21-3.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an%E4%B8%8Ebn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E5%88%86%E5%88%AB%E4%B8%BASn%E5%92%8CTn+%283n-2%29%E2%95%B1%282n+1%29)
∵an是Sn与2的等差中项∴2an=Sn+2(*)令n=1,得2a1=S1+2=a1+2∴a1=2由(*)得:2a(n+1)=S(n+1)+2两式相减,得:2a(n+1)-2an=a(n+1)即a(n
An=[2n/(3n+1)]BnAn-1=[2n/(3n+1)]Bn-1lim(n→∞)an/bn=lim(n→∞)[An-An-1]/[Bn-Bn-1]=lim(n→∞)[2n/(3n+1)][Bn
Sn=An²+Bn+C,{an}成等差数列的充要条件为C=0;S1=A+B+C=a1S(n-1)=A(n-1)²+B(n-1)+Can=Sn-S(n-1)=A(2n-1)+B已知a
由题意可得a1b1=S1T1=524=13,故a1=13b1.设等差数列{an}和{bn}的公差分别为d1 和d2,由S2T2=a1+a1+d 1b1+b1 +d&nbs
(1)S5=5a1+10d=5+10d=45,d=4,a3=1+2d=9.T3=b1+b2+b3=1+q+q^2=9-q,则q=-4或q=2.因为q>0,所以q=2.{an}的通项公式为:an=1+4
/>本题考察的是等差中项的概念.因为数列{an}是等差数列,因此:a1+a2+a3=(a1+a3)+a2=2a2+a2=3a2=12∴a2=4设该等差数列的公差为d,则:d=a2-a1=4-2=2因此
由AnBn=7n+45n+3,可设An=kn(7n+45)⇒an=An-An-1=14kn+38k,设Bn=kn(n-3)⇒bn=Bn-Bn-1=2kn+2k,所以a2n=28kn+38k,a2nbn
Ⅰ∵数列an是首项为16,公差为32的等差数列∴an=a1+(n-1)d=16+32(n-1)=32n-16Sn=(a1+an)n/2=(16+32n-16)n/2=16n²数列bn的前n项
∵{an}为等差数列,其前n项之和为Sn,∴S2n-1=(2n−1)(a1+a2n−1)2=(2n−1)×2an2=(2n-1)•an,同理可得,S′2n-1=(2n-1)•bn,∴anbn=S2n−
S(2n-1)=(A1+A(2n-1))×(2n-1)/2=(A1+A1+(2n-2)d)×(2n-1)/2=(A1+(n-1)d)×(2n-1)=An×(2n-1)同理T(2n-1)=Bn×(2n-
a3=a1+2d=7S11=11a1+11*10*d/2=11a1+55d=143{a1+2d=7,{a1+5d=13解得,a1=3,d=2an=a1+(n-1)d=2n+1bn=2^(2n+1)Tn
19/31An/Bn=[a1+(n-1)d]/[b1+(n-1)s]=2n/3n-1对比得到:a1=2d=4b1=8s=6a10/b10=38/62=19/31
{an}是等差数列,a2=a1+da3=a1+2d....an=a1+(n-1)da(2n-1)=a1+(2n-2)da1+a(2n-1)=2a1+(2n-2)d2an=2a1+2(n-1)d=2a1
证明:设等差数列{an}的首项为a1,公差为d,则Sn=na1+n(n−1)d2.bn=Snn=a1+n−12d.则bn+1−bn=a1+n2d−a1−n−12d=d2.∴数列{bn}是等差数列.
∵等差数列{an}{bn}的前n项和分别为Sn,Tn,∵SnTn=7nn+3,∴a5b5=s9T9=7×99+3=6312=214,故答案为:214
再问:额那个倒M是什么玩意儿,我们解数列都不用那个的再答:求和符号你可以理解成从第一个数加到第n个数……难道你不是高中……?再问:以前高一高二没认真听,所以不知道这是啥意思再答:你不用知道就是个表示形
n=1时,a1=S1=a+bn≥2时,Sn=a×n²+bnS(n-1)=a×(n-1)²+b两式相减得:an=Sn-S(n-1)=2a×n-a∴a(n-1)=2a×(n-1)-a∴
1.An/Bn=(7n+45)/(n+3)=(7n+7*3+24)/(n+3)=7+24/(n+3)An/Bn为整数,只需要24/(n+3)为整数,又n+3>3,则(n+3)=4,6,8,12或24得
答:等差数列An=1+2nBn=(An)^2-1=(An-1)(An+1)=2n(2n+2)=4n(n+1)=4n^2+4nSn=4*[(1^2+2^2+3^2+...n^2)+(1+2+3+...+
等差数列数列的性质a1+a[2n-1]=2an因为S[2n-1]=[(2n-1)(a1+a[2n-1])]/2=(2n-1)anT[2n-1]=[(2n-1)(b1+b[2n-1])]/2=(2n-1