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