Mathematics: Show that 2^2n-1 +1 is divisible by 3 for all n 1.

Show that 2^2n-1 +1 is divisible by 3 for all n 1.

shaeelynn3734

02.01.2022

The proof makes use of congruences as follows:

Step-by-step explanation:

We can prove this result using congruences module 3. First of all we shall show that

$2^{2n-1}\equiv 2 \pmod{3}$ for all $n\in \mathbb{N}$ . By induction we have

. For

we have $2^{4-1}=8\equiv 2 \pmod{3}$ Suppose that the statement is true for n=k

and let's prove that it is also true for n=k+1

. In fact, $2^{2(k+1)-1}=2^{2k-1+2}=2^{2k-1}2^{2}\equiv 2\cdot 2^{2}\equiv 8 \equiv 2 \pmod{3}$

Then induction we proved that $2^{2n-1}\equiv 2 \pmod{3}$ for all . Then

$2^{2n-1}+1\equiv 2+1\equiv 3\equiv 0 \pmod{3}$

From here we conclude that the expression $2^{2n-1}+1$ is divisible by 3.

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wolfgirl10000

12.03.2020

418.5

Step-by-step explanation:

318 * .75 =238.5

12 *12 =144

144 * 1.25= 180

i believe the answer is 418.5

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