設(shè)等差數(shù)列{an}與{bn}的前n項(xiàng)和分別為Sn和Tn,且Sn/Tn=(3n+2)/(2n+1),limn→∞an/bn=
設(shè)等差數(shù)列{an}與{bn}的前n項(xiàng)和分別為Sn和Tn,且Sn/Tn=(3n+2)/(2n+1),limn→∞an/bn=
數(shù)學(xué)人氣：225 ℃時(shí)間：2020-10-01 05:22:01
優(yōu)質(zhì)解答
設(shè)an=a1+(n-1)d1,bn=b1+(n-1)d2∴ Sn=na1+n(n-1)d1 /2Tn=nb1+n(n-1)d2 /2∴ lim (n-->∞)Sn/Tn=lim (n-->∞) [a1+(n-1)d1 /2]/[b1+(n-1)d2 /2]= d1/d2∵ Sn/Tn=(3n+2)/(2n+1)∴ lim (n-->∞)Sn/Tn=3/2即 d1/d2=3/2∴...
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