ruijie2004
春芽
共回答了17个问题采纳率:82.4% 举报
由条件知p=1/2,所以抛物线方程为x^2=y,即Sn=n^2=na1+dn(n-1)/2,所以d=2,a1=1,an=1+2n,bn=2^(1+2n)
cn=(1+2n)*2^(1+2n)
Tn=c1+c2+.+cn=3*2^3+5*2^5+.+(2n+1)*2^(2n+1) (1)
2^2Tn=3*2^5+5*2^7+.+(2n+1)*2^(2n+3) (2)
(2)-(1)=-3Tn=-[16-2^(2n+4)]/3-(2n+1)*2^(2n+3)
Tn=[16-2^(2n+4)]/9+(2n+1)*2^(2n+3)/3
(应该问的是cn的前n项和吧)
1年前
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