Let n ≥ 5 be an integer. If 9n-8 n-1=64 α and 6 n -5 n -1=25 β, then α-β is equal to:

Question

Let n ≥ 5 be an integer. If 9n-8 n-1=64 α and 6 n -5 n -1=25 β, then α-β is equal to: (A) 1+ n C2(8-5)+ n C3(82-52)+ ldots+ n Cn(8n-1-5n-. 1) (B) 1+ n C3(8-5)+ n C4(82-52)+ ldots+ n Cn(8n-2-5n-. 2) (C)  n C 3(8-5)+ n C 4(82-52)+ ldots+ n C n (8 n -2-5 n -2) (D)  n C4(8-5)+ n C5(82-52)+ ldots+ n Cn(8n-3-5n-3)

Tardigrade Admin · Accepted Answer

α=((1+8) n -8 n -1/64)= n C 2+ n C 3 8+ n C 4 82+ ldots β= n C2+ n C3 5+ n C4 52+ ldots option (3) will be the answer.

Q. 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:

$1 +^{n} C_{2} (8 - 5) +^{n} C_{3} (8^{2} - 5^{2}) + \dots +^{n} C_{n} (8^{n - 1} - 5^{n -}$ 1)

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

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

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

$α = \frac{( 1 + 8 ) ^{n} - 8 n - 1}{64} =^{n} C_{2} +^{n} C_{3} 8 +^{n} C_{4} 8^{2} + \dots$
$β =^{n} C_{2} +^{n} C_{3} 5 +^{n} C_{4} 5^{2} + \dots$
option (3) will be the answer.

Q. Let n≥5 be an integer. If 9n−8n−1=64α and 6n−5n−1=25β, then α−β is equal to:

1+nC2​(8−5)+nC3​(82−52)+…+nCn​(8n−1−5n− 1)

1+nC3​(8−5)+nC4​(82−52)+…+nCn​(8n−2−5n− 2)

nC3​(8−5)+nC4​(82−52)+…+nCn​(8n−2−5n−2)

nC4​(8−5)+nC5​(82−52)+…+nCn​(8n−3−5n−3)