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Solution
$$a_2 = 2*1 + 5$$
$$a_3 = 2*(2 + 5) + 5 = 2^2 + 5*2 + 5$$
$$a_4 = 2^3 + 5*2^2 + 5*2 + 5$$
...
$$a_{100} = 2^{99} + 5*(2^{98} + 2^{97} + ... + 1)$$
$$= 2^{99} + 5*1*(2^{99} - 1)/(2-1) = 2^{99} + 5*2^{99} - 5 = 6*2^{99} - 5$$
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