# Show that is divisible by 64, whenever n is a positive integer.

Question:

Show that $9^{n+1}-8 n-9$ is divisible by 64, whenever $n$ is a positive integer.

Solution:

In order to show that $9^{n+1}-8 n-9$ is divisible by 64 , it has to be proved that,

$9^{n+1}-8 n-9=64 k$, where $k$ is some natural number

By Binomial Theorem,

$(1+a)^{m}={ }^{m} C_{0}+{ }^{m} C_{1} a+{ }^{m} C_{2} a^{2}+\ldots+{ }^{m} C_{m} a^{m}$

For $a=8$ and $m=n+1$, we obtain

$(1+8)^{n+1}={ }^{n+1} C_{0}+{ }^{n+1} C_{1}(8)+{ }^{n+1} C_{2}(8)^{2}+\ldots+{ }^{n+1} C_{n+1}(8)^{n+1}$

$\Rightarrow 9^{n+1}=1+(n+1)(8)+8^{2}\left[{ }^{n+1} C_{2}+{ }^{n+1} C_{3} \times 8+\ldots+{ }^{n+1} C_{n+1}(8)^{n-1}\right]$

$\Rightarrow 9^{n+1}=9+8 n+64\left[{ }^{n+1} C_{2}+{ }^{n+1} C_{3} \times 8+\ldots+{ }^{n+1} C_{n+1}(8)^{n-1}\right]$

$\Rightarrow 9^{n+1}-8 n-9=64 k$, where $k={ }^{n+1} C_{2}+{ }^{n+1} C_{3} \times 8+\ldots+{ }^{n+1} C_{n+1}(8)^{n-1}$ is a natural number

Thus, $9^{n+1}-8 n-9$ is divisible by 64 , whenever $n$ is a positive integer.

Leave a comment

Click here to get exam-ready with eSaral