To make it easier to visualize I let $x = { 3 }^{ 1505 }.$ This means that $3x = { 3 }^{ 1506 }.$ Therefore, the given expression becomes $3x - x = 2x.$ Now, substitute our value of $x$ back in to obtain $(2)({ 3 }^{ 1505 }).$ However, since that is not in the choices, we factor out another $3$ from ${ 3 }^{ 1505 }$ like so:

$(2)(3)({ 3 }^{ 1504 }) = (6)({ 3 }^{ 1504 }) = \boxed { 6 \times { 3 }^{ 1504 } }$

$\quad { \quad 3 }^{ 1506 }-{ 3 }^{ 1605 }\\ =\quad { 3 }^{ 1605 }\times 3-{ 3 }^{ 1605 }\quad \quad \quad Take\quad "{ 3 }^{ 1605 }"\quad common\\ =\quad { 3 }^{ 1605 }(3-1)\\ =\quad { 3 }^{ 1605 }(2)=(2)(3){ 3 }^{ 1604 }\\ =\quad 6\times { 3 }^{ 1604 }$

Because this is a multiple choice, parity is a valid option. Notice how both $3^{1506}$ and $3^{1505}$ are odd. Their difference is even. Because of this, there will be a $2$ somewhere in the factorization. The only even option choice is $\boxed{6 \times 3^{1504}}$ .

Sol: 3^1506 - 3^1505

= 3^1505 × 3^1 - 3^1505

= 3^1505(3^1-1)

= 3^1505(2)

= 3×3^1504×2

= 6×3^1504

Here's the logic I used to originally make this problem;

First I took the base formula of $3^{n} - 3^{n-1}$ , where n is greater than or equal to 2. I started with the first three possible solutions

$3^{2} - 3^{1} = 9 - 3 = 6$

$3^{3} - 3^{2} = 27 - 9 = 18$

$3^{4} - 3^{3} = 81 - 27 = 54$

It was here that I saw that the answer to each of these was divisible by 3, so...

$6 / 3 = 2$

$18 / 3 = 6$

$54 / 3 = 18$

This one took me a bit longer to figure out. I saw that the last two were once again multiples of 3, but the 2 threw me off. Until I remembered that any number to the power of 0 is equal to 1. Thus...

$2 = 2 x 3^{0}$

$6 = 2 x 3^{1}$

$18 = 2 x 3^{2}$

As I had already divided the original answers by 3, all I had to do was multiply the above information by 3 to find the solution. So

$3^{n} - 3^{n-1} = 6 x 3^{n-2}$

To verify my work, I used my formula to calculate answers up to n = 10. I do apologize if the work and explanation is crude. I'm not mathematician, merely a guy that really likes math.

In hindsight though, I do believe that while I initially assumed n had to be >= to 2, I believe that it can be any integer. I did the math up to n = -9 and each time it worked out.

just take 3^1504 common from equation then it will become 3^1504(3^2-3^1) so,3^1504(9-3) so answer will be 6*3^1504

3^1506-3^1505= 3 (3^1505)-3^1505=(3^1505) (3-1)=2 (3^1505)=2 3 (3^1504)=6 (3^1504)

$\large 3^{1506}-3^{1505}$

$\large =3(3^{1505}-3^{1505}$

$\large =3^{1505}(3-1)$

$\large =2 \times 3^{1505}$

$\large =2(3)(3^{1504})$

$\large =6 \times 3^{1504}$

$3^{1506}-3^{1505}$ = $3^{1505}[3-1]$ = $3^{1505}[2]$ = $6[3^{1504}]$

3^1506 - 3^1505

=3^1505 ( 3 - 1 )

=3 * 3^1504 ( 2 )

=6 * 3^1504

I think it is a lot easier to factor out 3^1505 and write the answer as 3^1505 (3-1) = 3^1505 (2)

3^1506-3^1505=3 3^1505-3^1505=3^1505(3-1)=2 3^1505=6*3^1504

Take 3^1505 as common it implies 3^1506 - 3^1505=3^1505(3-1) =6×(3^1504)

Just the way I thought about it: 3^1506 is 3^1505 X 3, so factoring out the 3^1505 you get 2 x 3^1505, which can be converted into 6 x 3^1504

3^1506-3^1505=3^1505(3-1) =3^1504 * 3 * 2=6*3^1504

Taking out 3^1505 common we ve: 3^1505(3-1) =3^1505(2) . Again taking out 3 as common we ve 3 (3^1504) 2= 6*3^1504

3^1505x(3-1) =2x3^1505 =6x3^1504

3^1506-3^1505=(3^1504 3^2 )-(3^1504 3^1)=3^1504 (3^2 -3)=3^1504 (9-3)=3^1504*6

3^{1506} - 3^{1505}

=3^{1505}(3-1)

=3^{1505} * 2

=3 * 3^{1504} * 2

=3 * 2 * 3^{1504}

= 6 * 3^{1504}

3^1506 - 3^1505 = 3 * 3^1505 - 3^1505 = 2 * 3^1505 = 2 * 3 * 3^1504 = 6 * 3^1504

Answer: 6 * 3^1504

[(3^1505) x 3 - 3^1505] =[3^1505][3-1] =[3^1505][2] =[(3^1504) x 3][2] =6 x 3^1504

3^1505( 3-1) = 3^1505 * 2 3^1504 2 3 = 3^1504 *6

3^1506 - 3^1505 = 3^1505(3-1)

                                         =   3^1505 * 2


                                         =  3*2*3^1504


                                        = 6*3^1504

Sol: 3^1506 - 3^1505 = 3^1505 × 3^1 - 3^1505 = 3^1505(3^1-1) = 3^1505(2) = 3×3^1504×2 = 6×3^1504

use basic principle of algebra: a^m*a^n=a^(m+n)

just form a series of...

3^1 - 3 ^0 = 2 3^2 - 3^1 = 6 3^3 - 3^2 = 18 3^4- 3^3 = 54.....................................and so on...

check the options .... now,

2 can be written as...... 6 x 3^(-1) 6 can be written as ..... 6x 3^(0) 18 can be written as ... 6 x 3^(1)... so the only option that satisfies this form is ...

      6 x 3^(1504)....hahahahahahaha

Simply take 3^1505 common. Then the equation becomes as: => 3^1505(3^1-1) => 3^1505(2) Again take a 3^1 out of 3^1505 as: => 3^1504(3^1×2) =6×3^1504 is the required answer! ☺

Its simple..just take out the second term common and solve

The key is to take a common factor of 3^1505...

There are so many terms that can be taken as common. But according to the given options,

we will take 3^1504 common

then the solution will be 6×(3^1504)

$3^{1506}$ - $3^{1505}$ = ?

$3\times3^{1505}$ - $1\times3^{1505}$ = $2\times3^{1505}$

Rewriting $2\times3^{1505}$ = $2\times3\times 3^{1504}$ or $6\times3^{1504}$

taking 3^1505 common, we get the expression as 2(3^1505) which is the same as 6x3^1504

Actually the answer is 2 x 3^1505; but since it doesn't appear in the answer choices, the answer is 6 x 3^1504.

How to find it?

3^1506 - 3^1505

changes the 3^1506 into 3 x 3^1505: 3 x 3^1505 - 3^1505

then: 3^1505(3-1)

then 3^1505 x 2

then changes 3^1505 into 3 x 3^1504 3 x 3^1504 x 2

then: 6 x 3^1504

As we can see by playing around, any power of three subtract three to a power one less is just two times that last power of three.

3^2 - 3^1 = 6 = 2 * 3^1

and...

3^3 - 3^2 = 18 = 2 * 3^2

Generally:

3^(x) - 3^(x-1) = 2 * 3^(x-1)

Take out one factor of three on the RHS and we get one of the possible answers:

2 * 3^(x-1) = 6 * 3^(x-2)

Specifically:

3^1506 - 3^ 1505 = 6 * 3^1504

taking 3^1505 as a common factor we get 3^1505(3-1) which is equal to 3^1505 2=3^1504 3 2=6 3^1504

3^1506 - 3^1505= 3^1505(3-1)= 3^1505(2)= 3^1504 * 3 * 2= 6 * 3^1504

9 - 3 = 6, 6 = 2 x 3

27 - 9 = 18, 18 = 2 x 9

...

243 - 81 = 162, 162 = 2 x 81

Then ...

$3^{1506} - 3^{1505} = 2 x 3^{1505}$

$2 x 3^{1505} = 2 x 3 x 3^{1504} = 6 x 3^{1504}$

The same is 3^1505 so we can write 3^1505(3^1-3^0) =3^1505x2 or 6x3^1504. Simple man !

Using Tn = ar^(n-1)

3^1506 - 3^1505 = 3(3^1505)-3^1505 = 2(3^ 1505)= 2 (3^1054) x3 = 6 (3^1504)

3^1505(3-1)

=2*3^1505

=2 3 3^1504

=6*3^1504

2.3362448054767730667041342431036e+718

just take out the common. its just like child's play

3^{1506} – 3^{1505} = 3^{1505} (3 – 1) = 2 \times 3^{1505} = 2 \times 3 \times 3^{1504} = 6 \times 3^{1504}

3^1506 - 3^1505 = 3×3^1505 - 3^1505 = 3^1505(3 - 1) =2 × 3^1505