# How do you find the first four terms given a_1 = 5; a_n = 2a_(n-1) + 4?

Aug 2, 2016

$5 , 14 , 32 , 68$

#### Explanation:

$\left\{\begin{matrix}{a}_{1} = 5 \\ {a}_{n} = 2 {a}_{n - 1} + 4\end{matrix}\right.$

This formulation describes the sequence ${a}_{1} , {a}_{2} , {a}_{3} , \ldots$ recursively.

We find:

${a}_{1} = 5$

${a}_{2} = 2 {a}_{1} + 4 = 2 \left(5\right) + 4 = 14$

${a}_{3} = 2 {a}_{2} + 4 = 2 \left(14\right) + 4 = 32$

${a}_{4} = 2 {a}_{3} + 4 = 2 \left(32\right) + 4 = 68$

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Bonus

The general formula for a term of this sequence is described by the formula:

${a}_{n} = 9 \cdot {2}^{n - 1} - 4$

as we can verify with a proof by induction:

Base case:

$9 \cdot {2}^{\left(1\right) - 1} - 4 = 9 \cdot {2}^{0} - 4 = 9 \cdot 1 - 4 = 5 = {a}_{1}$

Induction step:

$9 \cdot {2}^{\left(n\right) - 1} - 4 = 2 \left(9 \cdot {2}^{\left(n - 1\right) - 1} - 4\right) + 4 = 2 {a}_{n - 1} + 4$