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∵數列是1,4,13,40121,…∴4=1+3^1,13=4+3^2,40=13+3^3121=40+3^4,…即:a[n+1]=a[n]+3^n可以用待定係數法,但由於比較簡單,我們直接兩邊减3^(n+1)/2=(3/2)3^n,得:a[n+1]-3^(n+1)/2=a[n]-3^n/2∵a[1]=1∴{a[n]-3…
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