已知数列{a n }满足a n+1 =2a n +3n 2 +4n+5,a 1 =1,求数列{a n }的通项公式.

2025-06-22 21:38:49
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回答1:

∵a n+1 =2a n +3n 2 +4n+5,
∴a n+1 +3(n+1) 2 +10(n+1)+18=2(a n +3n 2 +10n+18),
又∵a 1 +3+10+18=1+31=32,
∴数列{a n +3n 2 +10n+18}是以32为首项、2为公比的等比数列,
∴a n +3n 2 +10n+18=32×2 n-1 =2 n+4 ,
∴a n =2 n+4 -3n 2 -10n-18.