Sorry, sometimes the forum does not show properly the formatting (or maybe I am wrong).

It is NOT \(2n + 1\) but it is (if you notice very carefully) that 2 (in the first quantity) is \(2^{n+1}\)

Ok, yes the forum wasn't showing it right.

Thanks

I edit my answer then. The exponent in quantity A can be rewritten as \(2^{n+1} = 2n+2\). Thus, the exponent is always even and -1 to an even exponent becomes 1, so the answer would be C, not D. Sorry for bothering you but may you provide the OE?

Shouldnt the answer be "B" ?
becuase A is always -1 and B is always 1

Thank you now I get it

Carcass, you are still writing it (D)? I agree that it's (C).

ANswer IS D

You did right in considering the exponent as 2n+2. Since n is an integer, if for example you consider n=-1, then 2n+2 = 0. Then answer A would be equal to answer B. Thus you cannot get to a definite conclusion. Answer is D.

Regards

since n is a integer. if the n =1 then the value in A is -1 while in B is 1. if the n=2 then the both values are equal.
So the answer is D.

I believe the answer is C.

the question is (-1) to the power of 2 to the power of (n+1)

So to solve first we have to designate different values for n (integer) including +ve and -ve values.

if n = +1 => the last power will be (1+1). so it will be (-1) to the power of (2) to the power of (2) which is (-1) to the power of 4 and thus equals to 1.
Answer is C

if n = -1 => the last power will be (0). So it will be (-1) to the power of (2) to the power of (0), which is (-1) to the power of (0) which is also equal to 1.
Answer is C

if n = 2 => the last power will be (3). So it will be (-1) to the power of (2) to the power of (3), which is (-1) to the power of (6) which is also equal to 1.
Answer is C.

if n=-2 => the last power will be (-1). So it will be (-1) to the power of (2) to the power of (-1), which is (-1) to the power of (-2), which is synonymous to the 1/(-1)^2 which equals to 1 eventually.

Answer is still C.

I think the idea over here in this question is that we have two exponents, last of them is n+1, whatever value we put its going to be multiplied by (2) and thus becoming even, although it might be negative (the -ve sign in the exponent signals a reciprocal) the reciprocal of 1 is going to be 1. So since the value - or reciprocal- of (-1)^even it is always going to give us a positive signed 1.

Thus the answer MUST be C.

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Thanks!!!

Since its (-1)^(2^(n+1)) and n is an integer, if we take n to be say -2
the equation would become
(-1)^(2^(-1))
= (-1)^(1/2)
= sqrt(-1)
which is not defined hence answer is D

I found this problem easiest to considering some scenarios that might yield interesting results.

n=1 --> A = 1, B = 1 --> A=B
n=0 --> A = 1, B = 1 --> A=B
n=-1 --> A = -1, B = 1 --> A<B

Hence, the answer is D.

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