32 n+2-8 n-9, n ∈ N, is always divisible by

Question

32 n+2-8 n-9, n ∈ N, is always divisible by (A) 60 (B) 64 (C) 35 (D) 32

Tardigrade Admin · Accepted Answer

32 n+2-8 n-9=(32)n+1-8 n-9 =(1+8)n+1-8 n-9 =(1+ n+1 C1 ⋅ 8+ n+1 C2 ⋅ 82 ldots ldots)-8 n-9 = n +1 C2 ⋅ 82+ n+1 C3 ⋅ 83 ldots ldots ∴ 32 n+2-8 n-9 text is divisible by 64

Q. $3^{2 n + 2} - 8 n - 9, n \in N$ , is always divisible by

60

64

35

32

$3^{2 n + 2} - 8 n - 9 = (3^{2})^{n + 1} - 8 n - 9$
$= (1 + 8)^{n + 1} - 8 n - 9$
$= (1 +^{n + 1} C_{1} \cdot 8 +^{n + 1} C_{2} \cdot 8^{2} \dots\dots) - 8 n - 9$
$=^{n + 1} C_{2} \cdot 8^{2} +^{n + 1} C_{3} \cdot 8^{3} \dots\dots$
$∴ 3^{2 n + 2} - 8 n - 9 is divisible by 64$

Q. 32n+2−8n−9,n∈N, is always divisible by

60

64

35

32

32n+2−8n−9=(32)n+1−8n−9 =(1+8)n+1−8n−9 =(1+n+1C1​⋅8+n+1C2​⋅82……)−8n−9 =n+1C2​⋅82+n+1C3​⋅83…… ∴32n+2−8n−9 is divisible by 64

Q. $3^{2 n + 2} - 8 n - 9, n \in N$ , is always divisible by

$3^{2 n + 2} - 8 n - 9 = (3^{2})^{n + 1} - 8 n - 9$
$= (1 + 8)^{n + 1} - 8 n - 9$
$= (1 +^{n + 1} C_{1} \cdot 8 +^{n + 1} C_{2} \cdot 8^{2} \dots\dots) - 8 n - 9$
$=^{n + 1} C_{2} \cdot 8^{2} +^{n + 1} C_{3} \cdot 8^{3} \dots\dots$
$∴ 3^{2 n + 2} - 8 n - 9 is divisible by 64$