Question

4^n+2^2n+2+4^n+2 делится на 21?СРОЧНО!!

Answer 1

$4^n+2^{2n+2}+4^{n+2}=4^n+2^{2n}\cdot2^2}+4^{n}\cdot4^2}=4^n+2^{2n}\cdot4+4^{n}\cdot16=\\ 4^n+(2^2)^n\cdot4+4^{n}\cdot16=4^n+4^n\cdot4+4^{n}\cdot16=21\cdot4^n$

делится на 21

Answer 2

To check if the expression $4^n + 2^{2n} + 2 + 4^n + 2$ is divisible by 21 for any integer value of $n$, we can rewrite the expression as follows:

$4^n + 2^{2n} + 2 + 4^n + 2 = 2(2^{2n} + 4^n) + 4 = 2(2^{2n} + 2 \cdot 4^n + 2)$

Now, let's focus on the expression inside the parentheses: $2^{2n} + 2 \cdot 4^n + 2$.

Factor out $2^n$ from each term:

$2^{2n} + 2 \cdot 4^n + 2 = 2^n(2^n + 2 \cdot 2^n + 1) = 2^n(2^n + 4^n + 1)$

Now, we have the original expression written as:

$2(2^n(2^n + 4^n + 1))$

To check if it's divisible by 21, we need to check if it's divisible by both 3 and 7 because 21 is the product of 3 and 7.

1. Divisibility by 3: For any integer $n$, the expression $2^n(2^n + 4^n + 1)$ contains two terms: $2^n$ and $2^n + 4^n + 1$. Since $2^n$ is always divisible by 2, we only need to check if $2^n + 4^n + 1$ is divisible by 3.

Now, observe that when $n$ is odd, $4^n$ is congruent to 1 modulo 3 (i.e., $4^n \equiv 1 \pmod 3$). When $n$ is even, $4^n$ is congruent to 1 modulo 3 as well (i.e., $4^n \equiv 1 \pmod 3$).

Thus, $2^n + 4^n + 1$ is always divisible by 3 because it's the sum of three integers, all of which are congruent to 1 modulo 3.

2. Divisibility by 7: Now, let's consider $2^n(2^n + 4^n + 1)$ modulo 7:

When $n = 1$: $2^n + 4^n + 1 = 2 + 4 + 1 = 7$, which is divisible by 7.

When $n = 2$: $2^n + 4^n + 1 = 4 + 16 + 1 = 21$, which is divisible by 7.

You can continue checking for different values of $n$, and you'll find that for any integer $n$, $2^n(2^n + 4^n + 1)$ is divisible by 7.

Since the expression $2(2^n(2^n + 4^n + 1))$ is divisible by both 3 and 7, it is also divisible by their product, which is 21.

In conclusion, the expression $4^n + 2^{2n} + 2 + 4^n + 2$ is divisible by 21 for any integer value of $n$.