If n ∈ N, then 72 n+33 n-3 ⋅ 3n-1 is always divisible by

Question

If n ∈ N, then 72 n+33 n-3 ⋅ 3n-1 is always divisible by (A) 25 (B) 35 (C) 45 (D) None of these

Tardigrade Admin · Accepted Answer

Putting n=1 is 72 n+23 n-3 ⋅ 3n-1, we get 72.1+23.1-3 ⋅ 31-1=72 20 ⋅ 30 =49+1=50 (1) Also, for n=2 72.2+23.2-3 ⋅ 32-1 =2401+24=2425 (2) From Eq. (1) and (2), it is always divisible by 25 .

Q. If $n \in N$ , then $7^{2 n} + 3^{3 n - 3} \cdot 3^{n - 1}$ is always divisible by

25

35

45

None of these

Putting $n = 1$ is $7^{2 n} + 2^{3 n - 3} \cdot 3^{n - 1}$ , we get
$7^{2.1} + 2^{3.1 - 3} \cdot 3^{1 - 1} = 7^{2} 2^{0} \cdot 3^{0}$
$= 49 + 1 = 50 (1)$
Also, for $n = 2$
$7^{2.2} + 2^{3.2 - 3} \cdot 3^{2 - 1}$
$= 2401 + 24 = 2425 (2)$
From Eq. (1) and (2), it is always divisible by $25$ .

Q. If n∈N, then 72n+33n−3⋅3n−1 is always divisible by

25

35

45

None of these

Putting n=1 is 72n+23n−3⋅3n−1, we get 72.1+23.1−3⋅31−1=7220⋅30 =49+1=50(1) Also, for n=2 72.2+23.2−3⋅32−1 =2401+24=2425(2) From Eq. (1) and (2), it is always divisible by 25 .

Q. If $n \in N$ , then $7^{2 n} + 3^{3 n - 3} \cdot 3^{n - 1}$ is always divisible by

Putting $n = 1$ is $7^{2 n} + 2^{3 n - 3} \cdot 3^{n - 1}$ , we get
$7^{2.1} + 2^{3.1 - 3} \cdot 3^{1 - 1} = 7^{2} 2^{0} \cdot 3^{0}$
$= 49 + 1 = 50 (1)$
Also, for $n = 2$
$7^{2.2} + 2^{3.2 - 3} \cdot 3^{2 - 1}$
$= 2401 + 24 = 2425 (2)$
From Eq. (1) and (2), it is always divisible by $25$ .