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By substitution,
and so on, so that
[tex]g(n)=6.1^{n-1}g(1)=2.7\cdot6.1^{n-1}
By substitution,
and so on, so that
[tex]g(n)=6.1^{n-1}g(1)=2.7\cdot6.1^{n-1}