JEE Mains · Maths · STD 11 - 7. binomial theoram
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 \(1+{ }^{n} C_{2}(8-5)+{ }^{n} C_{3}\left(8^{2}-5^{2}\right)+\ldots+{ }^{n} C_{n}\left(8^{n-1}-5^{n-1}\right)\)
- B \(1+{ }^{n} C_{3}(8-5)+{ }^{n} C_{4}\left(8^{2}-5^{2}\right)+\ldots+{ }^{n} C_{n}\left(8^{n-2}-5^{n-2}\right)\)
- 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 & Solution
Correct Answer
(C) \({ }^{n} C_{3}(8-5)+{ }^{n} C_{4}\left(8^{2}-5^{2}\right)+\ldots+{ }^{n} C_{n}\left(8^{n-2}-5^{n-2}\right)\)
Step-by-step Solution
Detailed explanation
\(\alpha=\frac{(1+8)^{ n }-8 n -1}{64}={ }^{ n } C _{2}+{ }^{ n } C _{3} 8+{ }^{ n } C _{4} 8^{2}+\ldots\) \(\beta={ }^{n} C_{2}+{ }^{n} C_{3} 5+{ }^{n} C_{4} 5^{2}+\ldots\) option \((3)\) will be the answer.
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