1. Tardigrade
  2. Question
  3. Mathematics
  4. 32 n+2-8 n-9, n ∈ N, is always divisible by

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. $3^{2 n+2}-8 n-9, n \in N$, is always divisible by

Binomial Theorem

Solution:

$3^{2 n+2}-8 n-9=\left(3^2\right)^{n+1}-8 n-9 $
$=(1+8)^{n+1}-8 n-9$
$=\left(1+{ }^{n+1} C_1 \cdot 8+{ }^{n+1} C_2 \cdot 8^2 \ldots \ldots\right)-8 n-9$
$={ }^{ n +1} C_2 \cdot 8^2+{ }^{n+1} C_3 \cdot 8^3 \ldots \ldots $
$\therefore 3^{2 n+2}-8 n-9 \text { is divisible by } 64$