1、解:a3+a13=a1+a15,,又15(a1+a15)/2=225,故可得a1+a15=30=a3+a13,又a3=5,故
a13=30-5=25,故公差d=(a13-a3)/(13-3)=(25-5)/10=2,故a1=a3-2d=5-2*2=1,故
an=a1+(n-1)d=1+2(n-1)=2n-1.
2、解:Tn=b1+b2+b3+......+bn=(2^a1+2*1)+(2^a2+2*2)+......(2^an+2n)=
(2^a1+2^a2+......+2^an)+(2*1+2*2+......2n)=(2^a1+2^a2+......+2^an)+2(1+n)n/2=
(2^a1+2^a2+......+2^an)+n(n+1),又an=2n-1,即{an}是正奇数集,故a1=1,a2=3,a3=5,an=2n-1,
(2^a1+2^a2+......+2^an)=2^1+2^3+2^5+......2^2n-1为等比数列,公比为4,即求其和为
(1-4^n)/(1-4)=(4^n-1)/3 故Tn=(4^n-1)/3 +n(n+1)=(4^n-1)/3+n^2-1/3.
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