If x = -1 and if n is the sum of the first 404 prime numbers

since N is sum of even count of first 404 prime numbers , we can say that 403 are odd and 1 is even so sum will be odd of N
for the given expression
\(x + x^N + x^{N+1} + x^{N+2}\)
-1+(-1)^O + (-1)^0 +(-1)^ O ; where O is odd sum value of N
-1-1+1-1
-2
option B

Points to remember which would help in such questions.
a. \(x^{any number}\)=1 where x=1.
b. \(x^{any even number}\)=1, where x=-1.
c. \(x^{any odd number}\)=-1, where x=-1.
d. 2 is the only even prime number.

So the equation \(x + x^N + x^{N+1} + x^{N+2}\),
where x=-1 can
N = 2 + the sum of 403 odd prime numbers = even + odd = odd
N+1= 2 + the sum of 404 odd prime numbers = even + even = even
N+2 = 2 + the sum of 405 odd prime numbers = even + odd = odd
Can be rewritten as,
(-1) + (-1)^odd+(-1)^even+(-1)^odd. = (-1) + (-1) + 1 + (-1) = 1 – 3 = -2
The correct answer is -2.

When calculating the sum of the first 404 prime numbers, why aren't we able to use the sum = avg x count formula, which would make the sum an even number?

ex. sum= avg x 404 (anything times even = even)

 

­You can't use it because that formula requires that the numbers are evenly spaced in the set and 2-3-5-7-11 have different distances between them. 

You don't even need to calculate the sum though. This problem is a play on even and odd exponents and how it affects negative signs. If you recognize that only the first prime number "2" is even and every other prime number to infinity is odd, then you know that an even + odd = odd number. 

Since you know that the exponent will be ODD you know that the negative 1 will carry. When you add +1 to an odd number, it becomes an even number.... in that case the -1 will be squared and it turns into +1. 

With that info you plug it in and see that -1 + -1^(ODD) + -1^(EVEN) + -1^(ODD) = -1 - 1 + 1 - 1 = - 2­

If I take the approach that 2 + (3+5+7+11...). The numbers inside the parenthesis are all odd numbers. So won't their sum also be even? and 2 + even = even

Using that I get the answer as 0.

Can someone help where I am going wrong?

Ans (B)

Let's first check whether n, which is the sum of the first 404 prime numbers, is even or odd by figuring out the pattern in the first initial prime numbers.

Prime no. = 2
# of terms = 1
Sum = 2

Prime no. = 2,3
# of terms = 2
Sum = 5

Prime no. = 2,3,5
# of terms = 3
Sum = 10

Prime no. = 2,3,5,7
# of terms = 4
Sum = 17

So we notice that for odd no. of prime numbers (2,3,5), the sum is even (10) and vice versa.

Now applying this logic on the given equation, we have been given 404 prime numbers, which is even. Therefore, its sum N must be odd.

-> x + x^N + x^(N+1) + x^(N+2)
-> x + x^(odd) + x^(even) + x^(odd)
-> (-1) + (-1)^(odd) + (-1)^(even) + (-1)^(odd)
-> (-1) + (-1) + 1 + (-1)
-> -2
 ­