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

(C) ${ }^{n} C_{3}(8-5)+{ }^{n} C_{4}\left(8^{2}-5^{2}\right)+\ldots+{ }^{n} C_{n}\left(8^{n-2}-5^{n-2}\right)$

(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) $1+{ }^{n} C_{3}(8-5)+{ }^{n} C_{4}\left(8^{2}-5^{2}\right)+\ldots+{ }^{n} C_{n}\left(8^{n-2}-5^{n-2}\right)$

(D) ${ }^{n} C_{4}(8-5)+{ }^{n} C_{5}\left(8^{2}-5^{2}\right)+\ldots+{ }^{n} C_{n}\left(8^{n-3}-5^{n-3}\right)$

Binomial Theorem — JEE Main · 2022 · 29 Jun 2022 · Shift 2 Question with Solution

Let

n \geq 5

be an integer. If

9^{n} - 8 n - 1 = 64 α

and

6^{n} - 5 n - 1 = 25 β

, then

α - β

is equal to:

(A) $1 +^{n} C_{2} (8 - 5) +^{n} C_{3} (8^{2} - 5^{2}) + \dots +^{n} C_{n} (8^{n - 1} - 5^{n - 1})$
(B) $1 +^{n} C_{3} (8 - 5) +^{n} C_{4} (8^{2} - 5^{2}) + \dots +^{n} C_{n} (8^{n - 2} - 5^{n - 2})$
(C) $^{n} C_{3} (8 - 5) +^{n} C_{4} (8^{2} - 5^{2}) + \dots +^{n} C_{n} (8^{n - 2} - 5^{n - 2})$ ✓ Correct
(D) $^{n} C_{4} (8 - 5) +^{n} C_{5} (8^{2} - 5^{2}) + \dots +^{n} C_{n} (8^{n - 3} - 5^{n - 3})$

Answer: (C)

Using the binomial theorem:

9^{n} = (1 + 8)^{n} = 1 + n \cdot 8 +^{n} C_{2} \cdot 8^{2} +^{n} C_{3} \cdot 8^{3} + \dots

So:

α = \frac{9 ^{n} - 8 n - 1}{64} = \frac{9 ^{n} - 8 n - 1}{8 ^{2}} =^{n} C_{2} +^{n} C_{3} \cdot 8 +^{n} C_{4} \cdot 8^{2} + \dots

Similarly:

6^{n} = (1 + 5)^{n} = 1 + n \cdot 5 +^{n} C_{2} \cdot 5^{2} +^{n} C_{3} \cdot 5^{3} + \dots

β = \frac{6 ^{n} - 5 n - 1}{25} = \frac{6 ^{n} - 5 n - 1}{5 ^{2}} =^{n} C_{2} +^{n} C_{3} \cdot 5 +^{n} C_{4} \cdot 5^{2} + \dots

Therefore:

α - β =^{n} C_{3} (8 - 5) +^{n} C_{4} (8^{2} - 5^{2}) + \dots +^{n} C_{n} (8^{n - 2} - 5^{n - 2})

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Single correct option