If P(n): " 2^{2n}-1 is divisible by k for all | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the statement $P(n):$ " $2^{2n}-1$ is divisible by $k$ for all $n \in N$ ".We can write $2^{2n}-1$ as $(2^2)^n - 1 = 4^n - 1$.Using the bino

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Given the statement $P(n):$ " $2^{2n}-1$ is divisible by $k$ for all $n \in N$ ".We can write $2^{2n}-1$ as $(2^2)^n - 1 = 4^n - 1$.Using the binomial expansion,$4^n - 1 = (3+1)^n - 1$.Expanding $(3+1)^n$ using the binomial theorem: $(3+1)^n = 1 + n(3) + \frac{n(n-1)}{2}(3^2) + \dots + 3^n$.So,$4^n - 1 = (1 + 3n + 9\frac{n(n-1)}{2} + \dots + 3^n) - 1 = 3n + 9\frac{n(n-1)}{2} + \dots + 3^n$.This expression is clearly divisible by $3$ for all $n \in N$.Thus,the value of $k$ is $3$.

If $P(n):$ " $2^{2n}-1$ is divisible by $k$ for all $n \in N$ " is true,then the value of $k$ is:

Similar Questions

If $n \in N$,then ${x^{2n - 1}} + {y^{2n - 1}}$ is divisible by

If $11^{12}-11^2=k(5 \times 10^9+6 \times 10^9+33 \times 10^8+110 \times 10^7+\ldots+33)$,then $k=$

If $2 \cdot 4^{2n+1} + 3^{3n+1}$ is divisible by $k$ for all $n \in N$,then $k=$

If $k \in N$,then $3^{3k} - 26k - 1$ is divisible by

The remainder when $3^{37}$ is divided by $80$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module