解ï¼
(1)
n=1æ¶ï¼a1=S1=(1²+3Ã1)/2=2
nâ¥2æ¶ï¼Sn=(n²+3n)/2 S(n-1)=[(n-1)²+3(n-1)]/2
an=Sn-S(n-1)=(n²+3n)/2 -[(n-1)²+3(n-1)]/2=n+1
n=1æ¶ï¼a1=1+1=2ï¼åæ ·æ»¡è¶³
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¬å¼ä¸ºan=n+1
(2)
bn=anÃ2n=(n+1)Ã2n=2(n²+n)
Tn=b1+b2+...+bn
=2[(1²+2²+...+n²)+(1+2+...+n)]
=2[n(n+1)(2n+1)/6 +n(n+1)/2]
=2n(n+1)(n+2)/3
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