Search any question & find its solution
Question:
Answered & Verified by Expert
$2^{3 n}-7 n-1$ is divisible by
Options:
Solution:
1090 Upvotes
Verified Answer
The correct answer is:
49
Let $P(n)=2^{3 n}-7 n-1 \Rightarrow P(1)=0, P(2)=49$
$P(1)$ and $P(2)$ are divisible by 49 .
Let $P(k)=2^{3 k}-7 k-1=491$
$$
P(k+1)=2^{3 k+3}-7 k-8=8(49 I+7 k+1)-7 k-8
$$
$$
=49(8 I)+49 k=49 \lambda
$$
where, $\lambda=8 I+k$, which is an integer.
$P(1)$ and $P(2)$ are divisible by 49 .
Let $P(k)=2^{3 k}-7 k-1=491$
$$
P(k+1)=2^{3 k+3}-7 k-8=8(49 I+7 k+1)-7 k-8
$$
$$
=49(8 I)+49 k=49 \lambda
$$
where, $\lambda=8 I+k$, which is an integer.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.