Question

Let n ≥ be an integer. If 9ⁿ – 8n–1 = 64α and 6ⁿ – 5n = 25β, then α – β is ______.

पर्याय

• 1 + ⁿC₂(8 – 5) + ⁿC₃(8² – 5²) + ... + ⁿC_n(8^n–1 – 5^n–1)

• 1 + ⁿC₃(8 – 5) + ⁿC₄(8² – 5²) + ... + ⁿC_n(8^n–2 – 5^n–1)

• ⁿC₃(8 – 5) + ⁿC₄(8² – 5²) + ... + ⁿC_n(8^n–2 – 5^n–2)

• ⁿC₄(8 – 5) + ⁿC₅(8² – 5²) + ... + ⁿC_n(8^n–3 – 5^n–3)

Answer 1

Let n ≥ be an integer. If 9ⁿ – 8n–1 = 64α and 6ⁿ – 5n = 25β, then α – β is `underlinebb(""^nC_3(8 - 5) + ""^nC_4(8^2 - 5^2) + ... + ""^nC_n(8^(n-2) - 5^(n-2))`.

Explanation:

Given: 9ⁿ – 8n – 1 = 64α and 6ⁿ – 5n – 1 = 25β

⇒ So, α = `(9^n - 8n - 1)/64`

⇒ α = `((1 + 8)^n - 8n - 1)/64`

⇒ α = `(""^nC_0 + ""^nC_1 8 + ""^nC_2  8^2 + ...... + ""^nC_n  8^n - 8n - 1)/64`

⇒ α = `( 1 + 8n + ""^nC_2  8^2 + ...... + ""^nC_n  8^n - 8n - 1)/64`

⇒ α = `""^nC_2 + ""^nC_3   8 + .... + ""^nC_n  8^(n-2)`

And β = `(6^n - 5n - 1)/25`

⇒ β = `((1 + 5)^n - 5n - 1)/25`

⇒ β = `""^nC_2 + ""^nC_3  5 + ...... + ""^nC_n  5^(n-2)`

Now, `α - β  ""^nC_3 (8 - 5) + ""^nC_4(8^2 - 5^2) + ....... + ""^nC_n(8^(n-2) - 5^(n-2))`