Respuesta :

It is choice (A). If you tried it, you would see that it works.

Remark Start by noticing that each member of the sequence ends in a one. That means that x^n must always be the same number no matter what n is. (n is defined as a whole number > 0 ). Your choices are 3^n , 4^n , 5^n and 6^n.

Solution.

3^n try a few to see what happens when you raise 3 to the first 4 numbers.

3^1 = 3; 3^2 = 9; 3^3 = 27; 3^4 = 81 The last number is the only one that ends in a one. All the others are all over the place. C is not the answer.

Now try 4 4^1 = 4; 4^2 = 16; 4^3 = 64; 4^4 = 256 D is not the answer

How about 6 6^1 = 6; 6^2 = 36; 6^3 = 216; 6^4 = 1296 This is your second best answer. The problem is that when you take away 4, you wind up with a 2 and not a 1 as the last digit. Alas B is not the answer.

The only answer left is A. When you subtract 4 away from 5^n you get something ending in 1. Answer: A

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