Corrigé du 29 P. 22

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\[\begin{aligned} v_0 &= \frac{0+3}{0+1} = \frac{3}{1} = 3\;;& \\ v_1 &= \frac{1+3}{1+1} = \frac 4 2 = 2\;;& \\ v_2 &= \frac{2+3}{2+1} = \frac 5 3\;;& \\ v_3 &= \frac{3+3}{3+1} = \frac 6 4 = \frac 3 2\;;& \\ v_4 &= \frac{4+3}{4+1} = \frac 7 5.& \end{aligned}\]

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