(文)已知数列{an}中,a1=2 an=3an-1+4(n≥2),求an及Sn.
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(文)已知数列{an}中,a1=2 an=3an-1+4(n≥2),求an及Sn. |
答案
∵an=3an-1+4(n≥2),an+2=3(an-1+2), ∵a1+2=4≠0, ∴数列{an+2}是以4为首项,3为公比的等比数列. ∴an+2=4×3n-1,∴an=4×3n-1-2. ∴Sn=a1+a2+…+an =4×(30+31+32+…+3n-1)-2n =4×-2n =2×3n-2-2n. |
举一反三
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