after the (n+1)th is =0, and that the (n+1)th term itself is always =\BB{2n-1}\cdot\frac{1}{1}=\BB{2n-1}, enables us to find the value (either numerical or algebraical) of any nth Number of Bernoulli B2n-1, in terms of all the preceding ones, if we but know the values of B1, B3…B2n-3.