Question

\begin{cases}a(1)=20\\\\
a(n)=a(n-1)\cdot\dfrac{3}2
\end{cases}

Answer 1

To solve the recurrence relation you've provided, we have the following:

1. The initial condition: \( a(1) = 20 \)
2. The recurrence relation: \( a(n) = a(n-1) \cdot \frac{3}{2} \)

We can compute a few terms to understand the pattern:

- \( a(1) = 20 \)
- \( a(2) = a(1) \cdot \frac{3}{2} = 20 \cdot \frac{3}{2} = 30 \)
- \( a(3) = a(2) \cdot \frac{3}{2} = 30 \cdot \frac{3}{2} = 45 \)
- \( a(4) = a(3) \cdot \frac{3}{2} = 45 \cdot \frac{3}{2} = 67.5 \)

From this, we can see that each subsequent term is multiplied by \( \frac{3}{2} \) from the previous term.

To find a general formula for \( a(n) \), we can express \( a(n) \) in terms of the initial value and the product of the multiplicative factor over \( n-1 \) steps:

\[
a(n) = a(1) \cdot \left(\frac{3}{2}\right)^{n-1}
\]

Substituting \( a(1) = 20 \):

\[
a(n) = 20 \cdot \left(\frac{3}{2}\right)^{n-1}
\]

Thus, the general solution for the recurrence relation is:

\[
\boxed{a(n) = 20 \cdot \left(\frac{3}{2}\right)^{n-1}}
\]