Question

Using Binomial Theorem, show that 9^n+1-8n-9 is divisible by 64

Answer 1

To show that \( 9^{n+1} - 8n - 9 \) is divisible by 64, we first rewrite \( 9^{n+1} \) in a more manipulable form using the Binomial Theorem.

We start with the expression:

\[
9^{n+1} = (8 + 1)^{n+1}
\]

Using the Binomial Theorem, we expand \( (8 + 1)^{n+1} \):

\[
(8+1)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} 8^k 1^{(n+1-k)} = \sum_{k=0}^{n+1} \binom{n+1}{k} 8^k
\]

In this expansion, notice that the term \( k = 0 \) gives \( 1^{n+1} \):

\[
\text{Term for } k=0: \binom{n+1}{0} 8^0 = 1
\]

The term for \( k=1 \):

\[
\text{Term for } k=1: \binom{n+1}{1} 8^1 = (n+1) \cdot 8
\]

The remaining terms, for \( k \geq 2 \), will all contain \( 8^2 = 64 \), which means they will be divisible by 64.

Thus, we can write:

\[
9^{n+1} = 1 + (n+1) \cdot 8 + \sum_{k=2}^{n+1} \binom{n+1}{k} 8^k
\]

Now, we need to substitute this back into our original expression:

\[
9^{n+1} - 8n - 9
\]

Substituting in what we have for \( 9^{n+1} \), the expression becomes:

\[
1 + (n+1) \cdot 8 + \sum_{k=2}^{n+1} \binom{n+1}{k} 8^k - 8n - 9
\]

Simplifying this:

\[
= 1 + 8n + 8 - 8n - 9 + \sum_{k=2}^{n+1} \binom{n+1}{k} 8^k
\]

Combining like terms gives:

\[
= 0 + \sum_{k=2}^{n+1} \binom{n+1}{k} 8^k
\]

Now, observe that the summation \( \sum_{k=2}^{n+1} \binom{n+1}{k} 8^k \) includes all terms for \( k \geq 2 \), each of which contains a factor of \( 64 \) (since the smallest \( k=2 \) contributes \( 8^2=64 \)). Therefore, all terms in this sum are clearly divisible by 64.

Thus, we conclude that:

\[
9^{n+1} - 8n - 9
\]

is divisible by \( 64 \). Hence, we have successfully shown:

\[
9^{n+1} - 8n - 9 \equiv 0 \mod 64
\]

This completes the proof that \( 9^{n+1} - 8n - 9 \) is divisible by \( 64 \).