Question

Let $n \geq 5$ be an integer. If $9^{n}-8 n-1=64 \alpha$ and $6^{ n }-5 n -1=25 \beta$, then $\alpha-\beta$ is equal to:

• A. $1+{ }^{n} C_{2}(8-5)+{ }^{n} C_{3}\left(8^{2}-5^{2}\right)+\ldots+{ }^{n} C_{n}\left(8^{n-1}-5^{n-}\right.$ 1)
• B. $1+{ }^{n} C_{3}(8-5)+{ }^{n} C_{4}\left(8^{2}-5^{2}\right)+\ldots+{ }^{n} C_{n}\left(8^{n-2}-5^{n-}\right.$ 2)
• C. ${ }^{ n } C _{3}(8-5)+{ }^{ n } C _{4}\left(8^{2}-5^{2}\right)+\ldots+{ }^{ n } C _{ n }\left(8^{ n -2}-5^{ n -2}\right)$
• D. ${ }^{n} C_{4}(8-5)+{ }^{n} C_{5}\left(8^{2}-5^{2}\right)+\ldots+{ }^{n} C_{n}\left(8^{n-3}-5^{n-3}\right)$

Answer 1

Answer: (C)

$\alpha=\frac{(1+8)^{ n }-8 n -1}{64}={ }^{ n } C _{2}+{ }^{ n } C _{3} 8+{ }^{ n } C _{4} 8^{2}+\ldots$
$\beta={ }^{n} C_{2}+{ }^{n} C_{3} 5+{ }^{n} C_{4} 5^{2}+\ldots$
option (3) will be the answer.