Linear Recursion → Explicit Formula Technique

Quick “shift‑to‑homogeneous” solution

1. Move to a fixed point.

   Write the recursion as

   $$ a_{n+1}-\alpha = 2\bigl(a_{n}-\alpha\bigr), $$

   where the constant α\alpha satisfies $\alpha = 2\alpha + 7\Rightarrow\alpha=-7$.

2. Define the homogeneous part.

   Let $b_{n}=a_{n}+7$.

   Then b1=a1+7=10b_{1}=a_{1}+7=10 and

   $$ b_{n+1}=2b_{n}\quad\Longrightarrow\quad b_{n}=10\cdot 2^{,n-1}. $$

3. Return to $a_{n}$.

   $$ a_{n}=b_{n}-7 = 10\cdot 2^{,n-1}-7. $$

4. Apply the inequality $a_{n}>1000$.

   $$ 10\cdot2^{,n-1}-7>1000 ;\Longrightarrow; 2^{,n-1}>\frac{1007}{10}=100.7 $$

   Small powers of 2: $2^{6}=64,;2^{7}=128$.

   Thus $2^{,n-1}$ first exceeds $100.7$ at $n-1=7\Rightarrow n=\boxed{8}$