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Question 129

If $$a_1 = 1$$ and $$a_{n+1} = 2a_n +5$$, n=1,2,....,then $$a_{100}$$ is equal to:

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

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