Let $f (n) = 8^{n} - 3^{n}$ , if $n$ is odd natural number then $f (n)$ is divisible by

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $5$
$f (n) = 8^{n} - 3^{n}$
Since, $n$ is odd
for $n = 1$ , we get $f (1) = 8^{1} - 3^{1} = 5$
for $n = 3$ , we get $f (3) = 8^{3} - 3^{3} = 5 (97)$
for $n = 5$ , we get $f (3) = 8^{5} - 3^{5} = 5 (6505)$
Therefore, by induction we can say that $f (n)$ is divisible by $5$ for odd $n$ .
Ans: C

Suggest Corrections

Similar questions

Q. If

n \in N, f (n) = 37^{n + 2} + 16^{n + 1} + 30^{n}

, then

Let $f (n) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} n & n + 1 & n + 2^{n} P_{n} & ^{n + 1} P_{n + 1} & ^{n + 1} P_{n + 2}^{n} C_{n} & ^{n + 1} C_{n + 1} & ^{n + 1} C_{n + 2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Then, f(n) is divisible by

Q. Let

f (n) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} n & n + 1 & n + 2^{n} P_{n} & ^{n + 1} P_{n + 1} & ^{n + 2} P_{n + 2}^{n} C_{n} & ^{n + 1} C_{n + 1} & ^{n + 2} C_{n + 2} \end{matrix} ∣ ∣ ∣ ∣ ∣

then

f (n)

is divisible by:

Q. Consider a sequence

f (x), n ϵ I^{+} \cup {0}

defined by

f (0) = 0, f (n + 1) = \sqrt{6 + f (n)}

then

l i m n \to \infty f (n)

Q. Let

f (n) = 10^{n} + 3 \cdot 4^{n + 2} + 5; \forall n \in N

. The greatest value of the integer which divides

f (n); \forall n

, is

Join BYJU'S Learning Program

Let f(n)=8n−3n, if n is odd natural number then f(n) is divisible by

The correct option is D 5f(n)=8n−3nSince, n is oddfor n=1, we get f(1)=81−31=5for n=3, we get f(3)=83−33=5(97)for n=5, we get f(3)=85−35=5(6505)Therefore, by induction we can say that f(n) is divisible by 5 for odd n.Ans: C

Let $f (n) = 8^{n} - 3^{n}$ , if $n$ is odd natural number then $f (n)$ is divisible by